1
MSCE
PHYSICAL SCIENCE PASS-WORD
(FORM 3 &4) NEW EDITION
DAUD M.ADAM, Bed. (Science)(University Of Malawi, Chancellor College):
Physical Science/Maths Teacher, Ndirande Hill Secondary School, BT
Cell: O999 156 027, 0888 456 027, 0881 577975
Email: adauadam@yahoo.com , WhatsApp: 0888 456 027
©mufti 2016 production
All rights reserved. No part of this copy may be reproduced, stored in retrieval
system, or transmitted in any form or by any means without the prior permission
of the producer.
Acknowledgements:
Thanks to Almighty Allah.
Special thanks should go to Mr. K. A. Adini of LISS for his inspirational and
academic support. I also owe huge debt of gratitude to relatives, friends
(GodwinZimba) and colleagues whose encouragement provided impetus needed
in this kind of work.
Introduction:
This “PHYSICAL SCIENCE PASS-WORD” is aimed at providing teachers and
secondary school learners with science information that is in tandem with the
Malawi School Certificate of Education P/science syllabus.
It is rich in content with well spelt out objectives. Plenty of examples have been
given with quality diagrams to deepen your understanding of scientific concepts.
In addition, the copy provides Maneb questions with suggested solutions for the
past three years.The last topic nuclear physics has several challenging questions
and their suggested solutions.
Buy your “Fizo pass-word” copy and you will appreciate!
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TABLE OF CONTENTS:
1. Properties of Matter…………4-21
Objectives
Kinetic Theory of Matter
Temperature scales
Gas Laws
Pressure
Expansion & Contraction.
How the topic has been featured and
some suggested answers.
2. Elements & Chemical bonding…..22-45
Objectives
Atomic structure
Periodic table
Bond formation
Periodicity and trends
Sulfur, sulfuric acid and sulphates
How the topic has been featured and
some suggested answers.
3. Chemical Reactions 1………46-61
Objectives
Chemical reactions, equations and how
to balance them
Mole concepts
Stoichiometry
Empirical formula
Concentration
Titration
Heats of Reaction
Questions &answers at the end of the
next topic.
4. Chemical Reactions 2………61-77
objectives
Oxidation / reduction
Half/ full equations
Displacement reaction
Corrosion
Electrolysis
Electroplating
Acids &Bases
How the topic has been featured
&some suggested answers.
5. Force & Motion……………….77-92
Objectives
vectors and scalars
graphs
Newton’s Laws of Motion
Free Fall
Force and mass
How the topic has been featured &
some suggested answers.
6. Organic Chemistry…………92-103
Objectives
Organic compounds
alkanols, alkanoic acids, esters
Flow diagram
Questions &answers see next topic
3
7. Organic Chemistry 2………103-114
Objectives
Isomerism
polymerization
Categories of Plastics
Waste disposal &Management
How the topic has been featured
and some suggested answers.
8. Electricity & Magnetism 1……114-138
Objectives
Static electricity
Current electricity
Resistance
Electric power
Use of power at house hold level
Magnetism
Transformer
Questions and answers ;next topic
9. Electricity & Magnetism 2……139-146
Objectives
Band theory
Semi-conductors
How the topic has been featured
and some suggested answers
10. Oscillation&Waves…146-171
Objectives
Oscillation
characteristics of an oscillating
particle
Classification of waves
Properties of waves
Lens
Ray diagrams
Camera / human eye
Projector
How the topic has been featured and
some suggested answers
11. Nuclear physics…………172-193
Objectives
Structure of an atom
Radioactive isotopes
Radioactive decay
Detecting radiation
Decay curves and Half lives
Dangers &Safety measures for
handling radiation
Uses of nuclear radiation
Practice question and some
suggested answers
12. Reference……………………194
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1.PROPERTIES OF MATTER.
OBJECTIVES:
By the end of this chapter learners should be able to:
Describe the Kinetic theory of matter.
Explain the cause of gas pressure.
Explain the relationship between the speed of molecules and the temperature.
Explain the meaning of absolute temperature.
Convert temperature from one unit to the other.
State the evidence for molecular motion.
Discuss the gas laws.
Apply the gas laws in problem solving.
Explain how a manometer works.
Derive a formula for liquid pressure.
Relate liquid pressure to everyday activities.
Explain the application of expansion and contraction in our every day.
MATTER.
o Matter refers to anything that possesses mass and occupies space (has volume).
o Matter comprises tiny (minute) particles unseen by naked eyes.
o There are three types of particles namely, molecules, atoms and ions.
o A molecule is the smallest particle of a pure substance that can exist on its own (can
exist independently).
o Molecules in a substance are held together by Inter-Molecular Forces (IMF). The IMF is
also known as Van Deel Waal’s forces.
o Further study reveals presence of other particles within molecules. Theses are atoms.
o An atom is the smallest particle that gets involved in chemical reaction. During
formation of new things, molecules break down to form atoms and the atoms get
rearranged to give rise to new species.
o It is also defined as a basic unit of matter.
o Atoms are held together by inter-atomic force (IAF) or chemical bonds.
o The Inter-atomic forces are much stronger than the inter-molecular forces. I.e. the force
of attraction among atoms is stronger than the force of attraction among molecules.
Breaking of chemical bonds (IAF) entails occurrence of a chemical reaction while
breaking of IMF suggests change of states of matter.
BASIC ASSUMPTIONS OF KINETIC MOLECULAR THEORY OF GASES:
1. Gases contain tiny (submicroscopic) particles.
2. The distance between the molecules is larger compared to the size of the molecules.
3. Gas molecules have no force of attraction for each other.
5
4. Gas molecules move in a straight line and in all direction, colliding frequently with each
other and with the walls of the container.
KINETIC THEORY OF MATTER
It states that matter is made up of tiny particles which are constantly moving.
These particles attract each other when close.
The word ‘kinetic’ means moving.
Matter exists as solid, liquid or gas.
Properties of Solids:
Solids have a definite shape due to a close arrangement of particles.
Particles in solids do not move but just vibrate in the fixed positions. However, the
effects of such vibrations can be disastrous.
The IMF is very large in solids. This is why most solids have a high Mpt and Bpt.
They are incompressible due to absence of spaces among particles.
They do not flow (do not take shape of container) due to strong force of attraction (IMF)
that restricts particle movement.
Examples of solids include bricks, wall, metal ball, stones, chalk etc.
Arrangement of Particles in solids
Properties of Liquids:
Particles are slightly far apart.
Particles move by sliding over each other.
Liquids take the shape of the container in which they are poured.
Liquids have a less IMF as compared to IMF in solids.
Examples include Paraffin, oil etc
Particle arrangement in liquids
Properties of Gases:
Particles are far apart.
Particles move randomly in all direction. This random movement of particles is called Brownian
motion.
Gases do take the shape of the container. They spread to all parts of environment.
The IMF is almost negligible (almost zero). This makes the particles always on the move.
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Gas particles moving at high speed and randomly
MOLECULAR MOTION AND GAS PRESSSURE
o The continuous collisions of gas molecules with one another and with the walls of the container
cause pressure on the walls of the containing vessels.
o Thus, a gas pressure is the pressure due to a continuous heating of the gas molecules with the
walls of the container.
o An increase in temperature causes an increase in pressure. This is so, because gas particle gain
extra energy for movement.
TEMPERATURE SCALES:
Temperature refers to degree of hotness or coldness of an environment.
Temperature scales refer to the range of numbers measuring level of hotness or coldness. We
shall describe two temperature scales namely: Celsius scale and absolute temperature scale.
A. CELSIUS/CENTIGRADE SCALE:
On this scale, temperature is measured in degrees Celsius, °C.
The numbers are chosen specifically so that pure ice melts at 0°C and pure water boils at 100°C
under standard temperature and pressure.
This scale can have positive and negative values.
B. ABSOLUTE TEMPERATURE SCALE:
It also called absolute scale.
It is a scale that measures temperature in Kelvin.
It only registers positive values.
The lowest value on this scale is absolute zero.
Absolute temperature is the temperature measured or calculated on the absolute scale in
Kelvin.
Absolute zero is the temperature at which molecular motion stops or particles have a lowest
kinetic energy possible.
The absolute scale is also called thermodynamic scale.
CONVERSION OF UNITS
To change from degrees Celsius to absolute temperature you need to add 273.
Thus,
273
0
C
TT
k
.
Example: change the following to absolute temperature
a) 20°C
b) 0°C
7
c) -127°C
d) 77°C
a. T
k
= 20+273 = 293K
b. T
k
= 0+273 = 273K
c. T
k
= -127 + 273 = 146K
d. T
k
= 77+273 = 350K.
To change from absolute temperature to degrees Celsius just subtract 273.
273
0
k
TT
C
.
Change the following to degrees Celsius.
a. 500K
b. 200K.
a)
CTT
k
C
0
0
227273500273
b)
CTT
k
C
0
0
73273200273
Exercise
a. Convert the following temperatures to degrees Celsius
i. 120K
ii. 400K
iii. 0K
b. Convert the following temperatures to kelvins
i. 300⁰C
ii. -153 ⁰C
iii. 0 ⁰C
EVIDENCE FOR MOLECULAR MOTION
a) Occurrence of chemical reaction:
Exchange of atoms which takes place during chemical reaction implies movement.
If reacting particles were not moving they would not come to contact with each other and no
reaction would take place.
b) Diffusion:
This is the movement of particles from regions of higher concentration to regions of lower
concentration.
Fig below shows what happens when a jar with a dense brown gas, nitrogen dioxide is put
underneath a jar of air.
Moving particles of air and NO
2
spread themselves between the two jars.
Air
Cover. Five minutes after the cover is removed, air
and NO
2
mix up. Diffusion takes place.
NO
2
8
Diffusion also occurs in liquids where some colored crystals of soluble solids dissolve in water.
One can see the particles splitting off from the crystal and spread out in the water.
Examples of such kind of crystals are Copper sulphate and potassium permanganate.
Diffusion is faster in gases than in liquids.
Lighter particles diffuse faster and cover a larger distance than the heavier ones.
The rate of diffusion is affected by the following:
o Temperature.
o Mass (size) of the particle.
o Concentration.
o Surface area.
c) Brownian motion:
This, as already defined, is the random movement of particles. It acts as evidence of molecular
motion. Usually a microscope is used to trace a path moved by one particle of smoke enclosed in a
glass box. A pattern that clearly describes the random movement of the particles is drawn.
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
GAS LAWS
All gases have pressure. Pressure of gases is due to molecules hitting the sides of
container.
The magnitude (size) of the pressure is affected by the rate of collisions.
The interdependence of pressure, temperature and volume of gas was investigated in
the 17
th
and 18
th
century.
The relationship among these three quantities together with the laws of mechanics
forms the foundations of the Kinetic Theory of Matter.
CHARLES LAW.
It states that for a constant amount of gas the volume is directly proportional to the
absolute temperature if the pressure is constant.
Thus, an increase in volume leads to an increase in temperature
Symbolically,
KTVTV
where K is constant. Making K subject of the formula
we have
K
T
V
.
Dividing the volume by temperature one always get a constant number even if
conditions change to new ones. Thus,
2
2
1
1
T
V
T
V
or
2
2
1
1
V
T
V
T
SYMBOL
MEANING
V
1
Initial volume
T
1
Initial temperature
V
2
Final volume
9
T
2
Final temperature
Graphically,
V
Or V
-273 0 T(°C) T(K)
Example:
A helium sample at 25°C has a volume of 1.82 liters. If pressure and amount of gas are
unchanged determine what the volume will be when temperature is 50°C.
solution:
T
1
= 25+273 = 298K, T
2
= 50+273 = 323K, V
1
= 1.82L, V
2
=?
323298
82.1
2
V
LxV 97.1323
298
83.1
2
BOYLE’S LAW
It states that for a constant amount of gas, the pressure is inversely proportional to
volume if the temperature is constant. I.e.
V
P
1
for constant, T.
Graphically
V
P
- Boyle’s law with constant temperature becomes P
1
V
1
= P
2
V
2
e.g. A700ml sample
of gas at 500mmHg pressure is compressed at constant temperature until its final
pressure is 800mmHg. What is the final volume?
Initial state Final state
Pressure 500mmHg 800mmHg
Volume 700ml ?
P
1
, V
1
= P
2
, V
2
500mmHg × 700ml = 800mmHg × V
2
V
2
= 500mmHg × 700ml
800mmHg
V
2
= 438ml
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PRESSURE LAW
It states that for a constant amount of gas pressure is directly proportional to
the Kelvin temperature if the volume remains constant.
P α T for constant V. Graphically
P
T
E.g. A gas in a fixed container is at a pressure of 4atm and a temperature of 24
0
c
if the temperature is changed to 127
0
c, what would be the pressure?
P
1
= 4atm, P
2
=? T
1
= (27 + 273) K = 300K
T
2
= (127 + 273) K = 400K
2
2
1
1
T
P
T
P
k
P
k
atm
400300
4
2
atm
k
katmx
P 33.5
300
4004
2
IDEAL GAS LAWS
If all the gas laws are combined we would obtain a general form as follows:
PV = n RT where n R is constant.
The ideal gas law can also be written as
2
22
1
11
T
VP
T
VP
The standard atmospheric pressure and temperature (S. T. P.) are 1.01 × 10
5
Pascals
and 0
0
C (273K), 1atm = 1.01 × 10
5
pa = 760mmHg.
Ideal gas is the gas where pressure, volume and temperature behavior can be
completely described by the ideal gas equation. E.g. Hydrogen gas is contained in a
360cm
3
of glass and at standard temperature and pressure 0
0
C and 760mmHg.
Calculate the new pressure if it is carefully transferred into a glass bulb of volume
340cm
3
at 40
0
C?
V
1
= 360cm
3
, V
2
= 340cm
3
, T
1
= 0+273= 273K, T
2
=40+273= 313K, P
1
= 760mmHg,
Using:
2
22
1
11
T
PV
T
PV
11
313
340
273
760360
2
xPx
mmHg
x
xx
P 6.922
340273
313360760
2
Exercise
a.
6ml of fluorine gas F
2
at temperature of 27 °C is heated at constant pressure to
132°C. find its new volume Ans8.1ml
b.
20cm
3
of gas sample at a pressure of 1.5 atmospheres is transferred (at constant
temperature) to a container of volume 120cm
3
. Find its new volume
Ans 2.5 atmospheres
c.
0.5 liters of helium gas at STP (0°C, 1atm) is heated. Its new volume and pressure
become 1.4liters and 0.75atm respectively. Work out its new temperature
Ans 573.3K
d.
Table below shows values obtained through experiment on gas law.
Volume (mm
3
)
3.0
4.5
6.0
10.0
Pressure (Pa)
400
267
200
120
i. Plot a graph of pressure against volume
ii. Which gas law was being investigated in the experiment?
iii. Which two factors were made constant?
Solution
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ii. Boyle’s law
iii. Temperature and amount of gas
FACTORS THAT AFFECT GAS PRESSURE:
a) Number of moles:
The greater the number of molecules the bigger the pressure. This is so because there
are more molecules hitting the walls of the container.
b) Volume of Container:
o Decreasing the volume cause the gas molecules to bump into each sides and the lid
of container more often so that pressure increases.
o With constant number of molecules at constant temperature decreasing the
volume increases pressure.
c) Temperature:
With constant volume and constant number of molecules increasing the temperature
increases the pressure.
PRESSURE:
o It is defined as force per unit area.
0
100
200
300
400
500
0 2 4 6 8 10 12
Pressure (Pa)
Volume (mm3)
13
o Mathematically
)(
)(
2
mArea
NForce
P
o (N/m
2
) is Newton per square meter 1N/m
2
= 1 Pascal.
o E.g. Calculate pressure exerted by 200N force acting on the ground of area 40m
2
.
o
2
2
/5
40
200
mN
m
N
P
FACTORS THAT AFFECT PRESSURE
1. Force;
The higher the force the higher the value of pressure.
2. Area;
An increase in the area leads to a decrease in the value of pressure.
LIQUID PRESSURE
This is the pressure due to continuous hitting of liquid molecules with the walls of the
container.
The liquid pressure is affected by; height (depth) and density.
The higher the value of height or the density, the higher the value of liquid pressure.
PROPERTIES OF LIQUID PRESSURE
The liquid pressure is the same in all direction at the same depth
The liquid pressure increases with increase with depth (height)
The liquid pressure is perpendicular to the surface of the container
The liquid pressure increases with increase in density
EQUATION FOR LIQUID PRESSURE
It is given as: Liquid pressure= density x height x g where is g is the acceleration due to
gravity. Its value is 10N/Kg or 10m/s
2
.
Thus, P = dhg where P is the liquid pressure, d is density in Kg/m
3
, h is the height of the
liquid column in meters.
Examples
1. A container with water, density 1000Kg/m
3
supports a column of water of height 20
meters. Taking g = 10N/Kg calculate the value of liquid pressure at the bottom.
Solution: P = dhg = 1000 x 20 x 10 N/m
2
= 200000 Pa = 2 x 10
5
Pa.
2. find the liquid pressure in Pascals that is exerted by water at the bottom of a
container ( g = 10N/Kg)
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Solution
1g/cm
3
= 1000Kg/m
3
h = 6cm = 6/100cm = 0.06m
PadhgP 6001006.01000
Given Volume (Capacity) of the container and cross-sectional area:
area
volum e
height
Eg.A container with cross-sectional area of 15cm
2
is filled with 60cm
3
of paraffin of
density 0.8g/cm
3
. Calculate the pressure exerted by the paraffin at the bottom of the
container. Acceleration due to gravity, g = 10m/s
2
.
PadhgP
mcm
cm
cm
height
volume
h
mkgcmgd
3201004.0800
04.04
15
60
/80010008.0/8.0
2
3
33
see also Maneb (2013) #1c
DERIVING THE FORMULA FOR LIQUID PRESSURE:
Volume of Liquid = A x h
M = V x d
F = mg = v x d x g
F = A x h x d x g
dhg
A
Axdxhxg
A
F
P
hence
dhgP
USES OF LIQUID PRESSURE:
A. Hydraulic Systems:
o Hydraulic machines make use of the following properties:
- Liquids are incompressible
15
- The pressure of a trapped liquid is transmitted to all other parts of the
system.
o The fluid can be used to transfer forces from one place to the other.
o The basic principle is that the fluid transfers equal pressure throughout the
system.
o Examples of hydraulic machines include hydraulic brakes, hydraulic jacks,
hydraulic fork lifts etc.
Refer to the following diagram:
For example:
Work out:
I. the pressure piston A exerts on the fluid
II. The pressure the liquid exert on the piston B
III. The up-thrust (lift or the upward force) on the load.
Solutions:
I.
PamN
m
N
Ar ea
Force
P 250/250
2.0
50
2
2
.
II. The pressure the piston A exerts on the liquid = the pressure the liquid exerts on the
piston B = 250 Pa.
16
III.
NmxmNPxAF 500020/250
22
The hydraulic jacks are force multipliers because they increase the amount of force applied.
In the example above, force has being increased from 50 N to 5000N.
B. Construction of dams:
The dams are made thicker at the bottom in order to make them withstand huge
pressure at large depth.
C. Designing of outlet and in-let pipes:
The out-let pipes are located at the bottom since pressure is high hence liquid could
come out easily.
D. Location of water supply systems:
Reservoirs for water supply or hydraulic power stations are often made in hilly or
mountainous regions. This is to increase pressure.
E. Designing of sub-marines:
They are made of a tough and strong metallic and glass materials in order to prevent a
collapse due to high liquid pressure at the large depth of sea bodies.
LUNG PRESSURE:
This is the pressure exerted by the lungs on the air inside it.
It can be measured by a manometer e.g. a U-tube manometer
HOW A MONOMITER WORKS:
Air pressure from the lungs or a gas supply system forces the manometer liquid to rise
up in the open side of the u-tube until it remains steady when difference in level
balances with the lung pressure or gas pressure.
In general when end is connected to any supply of gas
Pressure of gas = atmospheric pressure + pressure due to liquid column
I.e. Gas pressure = atm pressure + hdg.
A manometer measures pressure difference.
The height difference shows the extra pressure that the gas supply has in addition to the
atmospheric pressure.
The extra pressure is called excess pressure of gas supply system.
To find pressure of gas supply one has to add the atmospheric pressure to its excess
pressure.
The following is an example:
17
I. Work out the pressure difference on the U-tube manometer.
Pressure difference = 140 mmHg – 100 mmHg = 40 mmHg.
II. If the atmospheric pressure is 760 mmHg work out the pressure of the gas.
Gas pressure = atmospheric pressure + gas pressure =
(760+40) mmHg= 800 mmHg.
ATMOSPHERIC PRESSURE:
This is the pressure air molecules exert on the ground. There is a blanket of various gas
particles in the atmosphere. It acts in all direction.
It can best be measured by barometer or a manometer.
It is highest at sea level and its value is about 100000N/m
2
= 100000Pa.
100000N/m
2
= 1 atm = 760 mmHg
The 100000 Pa is called the standard atmospheric pressure.
Atmospheric pressure decreases with height above sea level.
______________________________________________________________________________
EXPANSION AND CONTRACTION
Expansion is an increase in volume (size) while contraction is the decrease in volume or
size.
Metals expand differently when heated equally.
Consider a bi-metallic strip which is also called a compound bar. Two different metals
are juxtaposed side by side as shown below.
18
During heating, the metal that expands more bends towards the metal that expand less.
In the above strip copper expands more than iron hence when heated above room
temperature, the strip will bend as follows:
The metal that expands more also contracts more.
USES OF EXPANSION AND CONTRACTION IN OUR EVERYDAY DAY:
1. Construction of thermostats:
It is a device that regulates temperature in a circuit.
It is a compound bar made of metals that expand and contract differently.
2. Separating stuck tumblers:
This is done by passing hot water over the top tumbler. The top tumbler expands faster
than the inner one hence reducing the degree of stickiness.
3. Separating tightly screwed bottle tops:
It is done by pouring hot water on the bottle top. It expands before the glass does hence
removed with ease.
4. Shrink-fitting:
An example is the axle of gear wheel. There is a metal too big to enter into hole. It is
dipped in liquid nitrogen to cool it and make it contract. It becomes sizable to be
inserted into the wheel. Upon regaining normal temperature it expands to give a tight
fit.
5. Reverting of metals:
A hot revert (small metal) is placed into revert hole and its ends are hammered under
intensive heating. On cooling it contracts and pulls the plates being joined, together
6. Weathering:
It is the breaking down of rocks to form soil. It happens when a rock is subjected to
frequent contraction and expansion. It becomes loose, develops cracks and later
disintegrates into small fragments.
7. Construction of brick fences:
Gaps are left at intervals when constructing brick fences in order to provide room for
expansion and construction otherwise it may collapse.
19
Other uses include designing of fire alarms, and construction of railway lines.
8. Functioning of liquid-in –glass thermometer
Rapid expansion and contraction of the liquid assist in reading temperatures scales.
ANOMALOUS EXPANSION & CONTRACTION OF WATER:
Water expands and contracts abnormally.
Heating the water from 0°C to 4°C it contracts instead of expanding. From 4 to 0°C it
expands when factors suggest it has to contract.
a c
V
Volume (cm
3
)
b
4 Temperature (°C)
From a to b, volume of water is reducing and is minimum at b. Then volume increases as
temperature changes from 4°C onwards. It implies that water has largest density of
1g/cm
3
at 4°C.This abnormal expansion of water explains why pipes and bottles
containing water burst during cold days.
HOW THE TOPIC HAS BEEN FEATURED BY MANEB IN THE RECENT 4 YEARS:
2015
1. a state the pressure law
Answer: it states that for a fixed amount of gas at constant volume, the pressure is directly
proportional to the absolute temperature
b. The initial pressure of a fixed volume of a gas is 120mmHg. Calculate the final pressure in
mm Hg if the temperature of the gas is raised from
C27
to
C327
8c Explain how a bimetallic strip maintains the temperature of an electric iron at the required
level
mmHgP
P
T
P
T
P
mmHgP
KT
KT
240
300
600120
600300
120
120
600273327
30027327
2
2
2
2
1
1
1
2
1
20
The bimetallic strip acts as a thermostat. It is made of two metals with different rates of
expansion reverted together. When the iron overheats, the bimetallic strip bends towards the
metal that expands less there by breaking contacts (current no longer flows). When it cools,
the strip regains normal shape there by reconnecting the circuit. The heating cycle gets
repeated as long as the switch is on
2014
1
a. State any three uses of expansion of solids in everyday life
Answers:
Bimetallic strip, weathering of rocks, construction of brick fences, metal reverting etc
b. figure below is a diagram showing two similar tins A and B containing water of
density 1000kg/m
3
and petrol of density 800Kg/m
3
(i) Which liquid exerts a greater pressure at the bottom of the tin? Ans: Water
(ii) Give a reason for the answer in 1b(i) water has a higher density and the higher the
density the higher the pressure ( the other factor, height is the same)
(iii) Calculate the pressure at the bottom of tin B
8b Explain how a bimetallic strip maintains the temperature of an electric iron at the required
level
The bimetallic strip acts as a thermostat. It is made of two metals with different rates of
expansion reverted together. When the iron overheats, the bimetallic strip bends towards the
metal that expands less there by breaking contacts (current no longer flows). When it cools,
the strip regains normal shape there by reconnecting the circuit. The heating cycle gets
repeated as long as the switch is on
2013
1
a. Mention any two properties of liquid pressure
Pa
KgNmmkgdhgP
4000
/105.0/800
3
21
The liquid pressure is the same in all direction at the same depth
The liquid pressure increases with increase with depth (height)
The liquid pressure is perpendicular to the surface of the container
b. In terms of the kinetic theory of matter, explain why liquid evaporates
Answer: when liquid is heated, the motion of its particles increases due to gain
in kinetic energy. Further heating causes the particles to overcome the
attraction force among them (IMF) hence they start escaping from the surface
of the liquid as gas or vapor (evaporate).
c. A container with a cross-section area of 9cm
2
is filled with 36cm
3
of petrol.
Calculate the pressure exerted by the petrol at the bottom of the container.
(density of petrol = 0.7g/cm
3
and acceleration due to gravity (g) = 10m/s
s
)
PadhgP
mcm
cm
cm
height
volume
h
mkgcmgd
2801004.0700
04.04
9
36
/70010007.0/7.0
2
3
33
2012 a. State the kinetic theory of matter
Answer: It states that matter is made up of tiny particles which are always in motion
b Figure below shows an instrument used to measure pressure exerted by a gas
Calculate the pressure exerted by the gas if the atmospheric pressure is 765mm Hg
Answer: the excess pressure = 2.5cmHg = 25mm Hg
Gas pressure = excess pressure + atmospheric pressure
Hence Gas pressure = 25mm Hg + 765mm Hg = 790mm Hg
8cIn terms of kinetic theory of matter, explain why ice melts when put in the sun
Answer: Ice melts when put in the sun due to increase in heat energy which weakens the
IMF that holds the water particles together as the particles tend to vibrate more.
22
2.ELEMENTS AND CHEMICAL BONDING
Topic Objectives
By the end of this topic learners should be able to:
write the structure of an atom
Explain how elements differ from each other.
Analyze the arrangement of elements in the periodic table.
Predict the group and the period of an element given the atomic number.
Describe how atoms attain stability.
Distinguish between the ionic and covalent compound.
Represent bond formation with dot and cross diagrams.
Work out formula of compounds given valences.
Differentiate polar from non-polar covalent bonds.
Provide description of metallic bonds.
Describe properties of metals.
Draw relationship between properties and use of metals.
Describe physical and chemical properties of halogens.
Discuss uses of halogens.
State sources of sulphur.
Explain physical and chemical properties of sulfur.
Highlight uses of sulfur.
Discuss importance of sulfuric acid.
State uses of sulphates.
ATOMIC STRUCTURE
o An atom is the basic unit of matter. It gets involved during formation of new species
(Chemical reaction).
o Atoms contain other small particles referred to as ‘sub-atomic particles’. These are
protons, electrons and neutrons.
o Protons and neutrons together are referred to as nucleons.
o The number of protons and neutrons together gives the atomic mass number of an
element (nucleon number).
Protons.
o Are positively charged particles.
o Are found in the nucleus. (Nucleus is the central part of the atom).
23
o Have a mass of 1 amu. (1 atomic mass unit) which is the same as 1.6725 x 10
-27
kg.
Electrons
o Are negatively charged.
o They are located in the shells or energy levels.
o An electron has a mass of 9.1095 x 10
-31
kg.
o Are held in their positions by the positively charged nucleus.
Neutrons
o They are electrically neutral. Have a zero charge.
o Found in the nucleus.
o Have a mass of 1 amu or 1.6749 x 10
-27
kg.
Nuclide
In any atomic species of which the proton number is specified. i.e.
X
A
Z
A is the nucleon
number and Z is the atomic number.
Number of neutrons = Atomic mass – Atomic number = A –Z
E.g. given the following
Al
27
13
. State the number of
1. Protons
2. Electrons.
3. Neutrons.
Solution
1. 13
2. 13
3. 27 – 13 = 14
Draw the structure of the given element.
Ans:
Electrons
Nucleus
Shell (energy level)
AL
24
ELECTRON STRUCTURE, ARRANGEMENT OR CONFIGURATION
Electron configuration refers to the number and arrangement of electrons in the energy
levels.
Atoms with full outer energy levels are stable and show little chemical activity.
First electron shell holds up to 2 electrons.
The second shell holds a maximum of 8, so too with the third shell.
The forth shell holds up to 18.
E.g. Chlorine has the atomic number of 17. What is its electron configuration? It means it has 17
electrons hence its electron configuration is 2, 8, 7
THE PERIODIC TABLE OF ELEMENTS.
This is a tabular arrangement of elements listed in order of increasing atomic number.
The columns are called groups or families. There are 8 groups.
The horizontal rows are called periods or series and are 7 in total (if we consider the
whole periodic table).
The number of electrons in the outermost shell corresponds to group to which a
particular element belongs.
The number of shells corresponds to the period to which an element belongs.
E.g
ELEMENT
ELECTRON #
ELECTRON CONFIGURATION.
GROUP
PERIOD
Hydrogen
1
1
1
1
sodium
11
2,8,1
1
3
Nitrogen
7
2,5
5
2
Calcium
20
2,8,8,2
2
4
Argon
18
2,8,8
8
3
Chlorine
17
2,8,7
7
3
Carbon
6
2,4
4
2
The elements of the periodic table are categorized into three parts namely: metals,
metalloids and non-metals.
Some groups have special names.
Group 1: Alkali metals
Group 2: Alkaline Earth metals.
Group 7: Halogens.
Group 8: Noble gases, Inert gases or group 0.
25
METALS
They are found to the left of the periodic table.
Physical properties
They are good conductors of heat and electricity due to free moving electrons.
All are solids at room temperature (25°C) except mercury which is a liquid.
Strong under tension and compression.
Are malleable. (Capable of being shaped or bent or drawn out into shape). This is so
because the layers of ions can slide over each other. This makes them suitable in making
of iron sheets, cooking pots etc.
Are ductile. (Capable of being drawn into wires) hence used in making of electricity
transmission cables.
Are sonorous (make a ringing noise when hit.) hence used in making bells.
Have high densities due to close parking of their particles.
Are shiny when polished hence used in making jewelry and ornaments.
Have high melting and boiling points due to high attraction between the metal ions and
the sea of mobile electrons. It takes a lot of heat to break up the lattice except of alkali
metals.
Chemical properties.
Form positive metal ions (cations).
React with oxygen to form basic metal oxides.
NON-METALS:
They are found to the right of the periodic table. They are:
26
Are poor conductors of heat and electricity?
They have more varied physical properties than metals e.g. gases and liquids.
They are brittle (easily break when hit).
Have low melting and boiling points. Have low densities.
Form negative ions except Hydrogen. Their ions are called anions.
METALLOIDS:
They are found at the middle of the periodic table.
They have properties that fall between those of metals and the non-metals. Examples
include: boron, silicon, germanium etc.
Next part provides detailed description of the groups of the periodic table
ALKALI METALS (GROUP ONE).
Physical properties:
Are good conductors of electricity and heat.
Are soft metals.
Have shinny surfaces when freshly cut.
Are malleable (can be hammered into sheet).
Are ductile. (Can be molded into wires.)
Chemical properties:
They burn in air (oxygen) to produce oxides.
e.g.: Sodium + Oxygen to form Sodium Oxide.
ONaONa
22
They react with water vigorously to give an alkali solution and hydrogen gas.
Potassium + water to form potassium hydroxide and hydrogen gas.
)()()(
22
gHaqKOHOHsK
They react strongly with halogens to form salts (metal halides).
E.g. Lithium + Chlorine to form Lithium Chloride.
LiClClsLi
2
)(
Alkali metals have similar properties because their atoms all have 1 electron in their
outer shell.
Melting and Boiling point decrease down the group.
ELEMENT
MELTING POINT (°C)
BOIING POINT (°C)
Li
181
1342
Na
98
883
K
63
760
Rb
39
686
Cs
29
669
Reactivity of group one elements increases as we go down the group.
Group 1 metals become softer with low melting points as well as boiling point. This is so
because the electron in the outer most shell is loosely held by the nucleus so it easily lost.
27
Atomic radius increases down the group due to addition of shells.
ALKALINE EARTH METALS (GROUP 2).
Physical properties:
Are harder than those in group 1.
Are good conductors of heat and electricity.
Have high densities.
Generally, melting and boiling points decrease down the group.
ELEMENT
MELTING POINT (°C)
BOILING POINT (°C)
Be
1278
2970
Mg
649
1107
Ca
839
1484
Chemical properties:
They burn in oxygen to form oxide and a bright flame.
MgOOMg
2
They react with water but so much less vigorously than the group 1.
e.g.
222
)( HOHCaOHCa
They are less reactive because it is more difficult to lose two electrons than one because
loss of electrons requires use of energy.
Reactivity increase down the group.
______________________________________________________________________________
TRANSITION METALS:
Are harder and stronger.
Have high densities.
Are good conductors of heat and electricity
Many are used in making metal alloys (mixture of metals).
Are less reactive metals.
Form a range of brightly colored compounds.
They do not react (corrode) so quickly with oxygen. Most of them form ions with a plus
two charge.
Examples are: Copper, iron, zinc, lead, chromium, silver, gold, manganese, cobalt, nickel,
cadmium, mercury etc.
HALOGENS (GROUP 7):
These are the most reactive nonmetals.
Their reactivity increases up the group. This is so because the incoming electron is being
more strongly attracted into the outer energy level of the small atom.
28
Are more reactive because their atoms are just one short of full outer shell,
They have similar properties because their atoms have 7 electrons in their outer most
shell.
Physical properties of Halogens:
They are colored and they darken going down the group.
They exist as diatomic molecules because two atoms can gain full shells by sharing
electrons with each other.
Their boiling points and melting points increase going down the group because of the
increase in the force of attraction between the molecules
ELEMENT
FORMULA
M.Pt (°C)
B.Pt (°C)
Fluorine
F
2
-220
-187
Chlorine
Cl
2
-101
-35
Bromine
Br
2
-7.2
59
Iodine
I
2
113
188
.
They show a gradual change in physical state from a gas (Cl
2
) through a liquid (Br
2
) to a
solid (I
2
) at room temperature.
Their atomic radii as well as ionic radii increase down the group. This is due to addition
of shells.
Chemical properties of Halogens:
Are poisonous nonmetals.
They form molecular compounds with other nonmetallic elements. e.g. CCl
4
, PF
3
, S
2
Cl
3
etc.
They react with hydrogen to form halides. E.g. HF, HCl, HI etc.
They react with metals to form salts hence called halogens- salt formers.
NaClClsNa
2
)(
They undergo displacement reaction with each other. A more reactive metal displaces a
less reactive metal from solutions. e.g.: Molecular chlorine can displace bromide and
iodide ions in solutions. In general, Halogen upper in the group tend to displace those
below.
They react with oxygen.
They undergo inter –halogen reaction.
e.g :
ClFClF 2
22
USES OF HALOGENS:
a. Fluorine, F
2
It is used in the form of fluoride in drinking water and tooth paste because it reduces
tooth decay by hardening the enamel on the teeth.
29
It is used to produce poly tetra fluoro ethylene, a polymer better known as Teflon.
used to make commercial refrigerant gas which contains CCl
2
F
2
etc.
It is used as a volatile component in aerosol cans.
b. Chlorine. Cl
2
It is a poisonous dense green gas.
Used to make PVC pipes
It is used in house hold bleaches.
Used to kill microorganisms in drinking water and swimming pools.
Used to make organic solvents e.g. CCl
4
and chloroform for removal of grease.
Used in the manufacture of insecticides e.g. sodium chlorate.
Used in the production of hydrochloric acid.
Used in sewage treatment
Used in aesthetics, antiseptics e.g. dettol and califlavin.
c. Bromine. Br
2
It is a caustic and toxic brown volatile liquid.
Used to make disinfectants, medicines and fire retardants.
Used in photographic film.
Used in manufacturing of 1,2, dibromoethane, C
2
H
4
Br
2
which is added to tetra ethyl
lead used as anti-knock in petrol engines where undesirable accumulation of lead is
prevented.
d. Iodine, I
2
It is a shiny black solid which sublimes to form a violet vapor on gentle heating.
Used in medicine as a germicide to kill bacteria in skin cuts and to flex muscles.
Used as a photographic chemical.
Used in making of dyes.
Used in testing of starch.
GROUP 8 (NOBLE GASES):
The members include helium, neon, argon, krypton, xenon and Radon.
Are non metals.
Are colorless gases which occur naturally in air.
They normally don’t react with anything. This is attributed to the fact that their outer
most shell is completely filled with electrons. Other elements do react in order to attain
a fully- electron filled outer shell.
BOND FORMATION:
An atom other than hydrogen tends to form bonds until it is surrounded by eight
valence electrons. This is called Octet rule. (Oct means eight).
30
The valence of an atom is the number of electrons gained, lost or shared in bonding
with one or more atoms. The electrons involved in bonding are the ones found in the
outer most shell which is called the valence shell.
How compounds or bonds are formed.
If the valence shell of an atom is full, the atom is said to be stable. By reacting with each
other, atoms obtain full outer shells and therefore achieving stability.
Atoms gain stability ( attain a full outer shell) by:
a. Loosing electrons.
b. Gaining electrons.
c. Sharing electrons.
Elements (metals) with 1, 2 or 3 electrons in the outer most shell tend to lose them in
order to attain the inert gas configuration. Their outer most shell will be the next lower
one.
Elements (non-metals) with 5, 6, 7 electrons in the outer most shell tend to gain
electrons in order to attain stability.
Element with 4 electrons on their outer most shell generally share electrons.
Dot and Cross Diagram:
This is a simple way of showing bond formation. A dot and a cross are used to represent
electrons contributed by each atom during bond formation.
Types of Bonds:
There are three types of chemical bonds namely:
1. Ionic bond.
2. Covalent bond.
3. Metallic bond.
1) Ionic bond.
It is also called electrovalent bond.
It is a bond formed due to loss and gain of valence electrons.
It involves a metal and a non-metal.
The bond involves ions of opposite charge which are held together by
electrostatic force. e.g.: Na
+
and Cl
-
ions.
The compounds formed by these cations (positive ions) and anions (negative
ions) are neutral.
The compounds with ionic bonding are called ‘ionic compounds’ or
‘electrovalent compounds.’
E.g. Sodium loses an electron to become Na
+
and chlorine gains the lost electron
to become Cl
-
. The compound formed will be sodium chloride with formula NaCl.
Consider Calcium and fluorine. Calcium is in group 2 hence it loses two electrons
to become Ca
+2
. The fluorine atom gains a single electron to become F
-
.
However, there is a single electron released from calcium atom that is remaining
31
hence there is a need of another fluorine atom to accept it. Thus, two fluorine
atoms are needed to combine with one atom of calcium. The formula of the
calcium fluoride is CaF
2
The dot and cross diagram of NaClandCaF
2
will be as follows:
NaCl
CaF
2
To work out formula of an ionic compound one just has to follow the following steps:
Write down the name of the ionic compound.
Write down the symbol for its ions.
The compound must have no overall charge, so balance the positive and negative
charges by changing the number of ions involved. The total charge has to be equal to
zero.
Write down the formula without the charges.
Exercises on Dot and cross diagrams
Draw dot and cross diagrams for the ionic compounds formed from the given elements.
(Include the name and formula of the compounds)
a. Potassium and oxygen
32
Name; Potassium oxide formula
OK
2
b. Magnesium and chlorine
Name: Magnesium chloride formula
2
MgCl
c. Aluminium and oxygen
Name:Aluminium oxide formula
32
OAl
Listed below are some commonly used ions.
Take note that the charge is equal to the number of electrons lost or gained which is the
valency. There will be a positive on the ion if it loses electrons and a negative if it gains
electrons.
CATIONS (POSITIVE IONS):
1
2
3
Hydrogen, H
+
Lithium, Li
+
Potassium, K
+
Sodium, Na
+
Ammonium, NH
4
+
Silver, Ag
+
Hydronium, H
3
O
+
Barium, Ba
2+
Iron (II), Fe
2+
Lead (II), Pb
2+
Zinc, Zn
2+
Calcium, Ca
2+
Copper (II), Cu
2+
Aluminium, Al
3+
Iron (III), Fe
3+
Chromium, Cr
3+
33
ANIONS (NEGATIVE IONS):
1
2
3
Fluoride, F
-
Chloride, Cl
-
Bromide, Br
-
Iodide, I
-
Hydroxide, OH
-
Nitrate, NO
3
-
Hydrogen carbonate,
HCO
3
-
Permanganate MnO
4
-
Cyanide, CN
-
Acetate C
2
H
3
O
2
-
or
CH
3
COO
-
Carbonate, CO
3
2-
Sulphate SO
4
2-
Sufide , S
2-
Oxide, O
2-
Dichromate, C
2
O
7
2-
Hydrogen phosphate, HPO
4
2-
Phosphate, PO
4
3-
Physical Properties of Ionic Compounds:
Are usually solids at room temperature with high melting points due to strong
electrostatic forces of attraction holding the crystals together. A lot of energy is needed
to separate the ions and melt the substance.
Are usually hard substances.
Do not conduct electricity when in solid form because their ions are not free to move.
Are soluble in water because water molecules are able to bond with the positive and
negative ions which cause the lattice to break up and hence keep the ions apart.
They conduct electricity in molten or aqueous form since the ions are mobile.
Do not dissolve in organic solvents such as petrol, carbon tetra chloride.
2) Covalent Bond (Molecular bond):
This is the bond formed by sharing of electrons in the outer energy levels.
It occurs between non-metal atoms.
Compounds are called covalent compounds or molecular compounds. e.g.,
methane, water, sugar and ammonia.
Covalent compounds have strong intra-molecular forces but weak inter-
molecular forces.
However, inter-molecular forces increase with an increasing size of the molecule.
A dot and a cross in the inter-section or a dash ( ) represents a shared
(bonded) Pair of electrons.
Types of Covalent bonds:
If one looks at the number of shared electrons then the covalent bonds will be categorized as:
I. Single Covalent bond.
It normally involves sharing of a pair of electrons.
e.g.: bonding in chlorine molecule, fluorine molecule etc.
34
Cl Cl
II. Double- Covalent bond
It involves sharing of 2 pairs of electrons. Each atom contributes 2 electrons for
sharing.
E.g.: bonding in carbon dioxide, bonding in an oxygen molecule. In CO
2
its
O C O
III. Triple Covalent bond.
Atoms in this type of bonding share3 pairs of electrons.
E.g.: bonding in Nitrogen and acetylene.
NN
The way elements share electrons can also give rise to two types of covalent bonds which are:
a) Polar covalent bond.
b) Non-polar covalent bond.
a. Polar covalent bond:
This is a covalent bond in which electron are not shared equally.
It happens because of differences in the electro negativity values.
Electro negativity is the ability of an atom in a covalent bond to attract the
bonding electron.
For example Fluorine. This is more electronegative than hydrogen hence in the
formation of hydrogen Fluoride molecule, sharing of electrons is not equal
because H and F are different atom with different electro negativity. This makes
35
the electrons spend more time in vicinity of one atom. In this case, electrons will
spend more time in the region of fluorine.
The type of bond formed in this way is called polar covalent bond or simply
polar bond.
The compounds with polar bonds their molecules have one end positively
partially charged and the other end negatively slightly charged. Thus, the bond in
polar compounds has some partial ionic characters.
In the periodic table, the electro negativity decreases as one goes down a group
but it increases across the period from left to right.
The opposite of the electro negativity is electro positivity which is the ability of
an atom to lose its outer most electrons in order to achieve stability. Electro
positivity increases down a group. Potassium is the most electro positive
element.
Properties of covalent compounds:
Are usually gases, liquids or solids of low melting points and boiling points.
The melting points are low because of the weak Van der Waals forces of attraction.
Giant molecular substances have higher meting points because the whole structure is
held together by strong covalent bonds with the molecules.
Generally they do not conduct electricity since when molten or dissolved in water they
do not ionize save for HCl and some few acids. The HCl, when dissolved in water, it
ionizes to produce H
+
and Cl
-
.
In general, they do not dissolve in water unless they react with it to form ion s.
However, water is an excellent solvent and can interact with and dissolve some covalent
molecules e.g. Sugars.
They dissolve in organic solvent such as petrol, carbon tetra chloride. etc.
Examples of covalent compounds include sugar, ammonia, carbon dioxide, water etc
DIFFERENCES BETWEEN IONIC AND COVALENT BONDS:
When salt (ionic) e.g. sodium chloride and sugar (covalent) are heated equally, sugar takes little
time to melt than the salt.
I. This indicates that ionic bonds are stronger than the covalent bonds.
II. Ionic bonds are formed by loosing or gaining valence electrons while the covalent bonds
are formed by sharing of electrons.
III. Ionic bonds are because of electrostatic forces while covalent bonds are due to
intermolecular forces.
IV. Ionic bonds occur between a metal and a non-metal while covalent bond involves non-
metals.
DIFFERENCES BETWEEN IONIC COMPOUNDS AND COVALENT COMPOUNDS:
36
a. Ionic compounds are generally soluble in water while the covalent (molecular)
compounds are insoluble in water but in organic compounds like paraffin etc.
b. Ionic compounds have high melting and boiling points hence need high heat energy to
melt while covalent compounds have low melting and boiling points hence melt at a
slight heating.
c. Ionic compounds conduct electricity in aqueous or liquid form while covalent
compounds do not conduct electricity in either state except some few acids.
3) Metallic bond.
Atoms in a metal are packed tightly together in a regular pattern.
The tight parking causes outer electrons to be separated from the atoms and the result is
a lattice of ions in a ‘sea’ of electrons.
The great cohesive force resulting from delocalization is responsible for the strength of
metals. The mobility of the delocalized electrons is responsible for the strength of
metals. The mobility of the delocalized electrons account for metals being good
conductors of heat and electricity.
Ions of metals in a sea of delocalized electrons
Metallic bonding is therefore defined as the electrostatic attraction binding the positive
ions of a solid metal together by means of a sea of delocalized electrons.
Examples are the bonding in Lithium metal, copper metal etc.
In Summary refer to the following Flow diagram:
37
______________________________________________________________________________
Periodicity of properties of elements in the periodic table:
Periodicity is the recurrence of similar properties at regular intervals when elements are
arranged in increasing atomic number order.
Examples of these periodic properties are: Atomic radius, atomic volume, density,
electro negativity, electro positivity, nuclear charge, effective nuclear charge.
Atomic radius.
o It is half the distance between the two nuclei in two adjacent atoms.
o It increases down the group due to addition of shells.
o It decreases across the periods due to an increase in nuclear charge which causes the
atom to shrink due to the pulling effect of the nucleus.
o Thus the atomic radius is affected by two factors namely: number of shells/ energy
levels and the nuclear charge (atomic number).
Nuclear charge.
o It is the number of protons in the nucleus. For example the nuclear charge of He is +2,
Na is +11 Cl is +17 etc.
38
o It increases across the period and down the groups.
o It affects the atomic radius.
The effective nuclear charge.
o It can be defined as the portion of the nuclear charge that is experienced by the highest
energy level electrons.
o Effective nuclear charge is always equal to the number of the outer most electrons. For
instance the effective nuclear charge of potassium is +1, for magnesium is +2 and for
Oxygen is +6.
Melting and boiling points.
o They increase to the middle of the periodic table and then decrease again. They are
lowest on the right.
Trends in the periodic table:
Number of outermost shell electrons increase by 1 each time like the group number.
Elements go from metals through metalloids to non-metals.
Metal atoms have fewer outer shell electrons and lose them during chemical reaction to
achieve full shell. Non –metals react by gaining or sharing electrons.
The melting and boiling points increase to the middle of the period then decrease again.
They are lowest on the right.
All elements except the noble gasses react with oxygen to form oxides.
Reactivity decreases across the period reaching the lowest at group 4 and then
increases again reaching highest at group 7.
All elements react with oxygen to form oxides except the group eight.
SULPHUR:
It is a yellow non-metallic tasteless and order-less, naturally occurring element.
It is the sixth most abundant element in the earth’s crust
It is in group 6 and period 3 of the periodic table.
Sources of sulfur:
From underground sulfur beds. This is the main source.
It is extracted from the underground by a process called Frasch process. In this process,
superheated water under high pressure is pumped down to melt the sulfur. Compressed
air is then forced down the inner pipe. Liquid sulfur is mixed with air forms an emulsion
that is less dense than water and therefore rise to the surface as it is forced up through
a middle pipe. Refer to the figure below:
39
From crude oil , natural gas and some organic compounds in the form of hydrogen
sulfide, H
2
S and sulfur dioxide,SO
2
.
Found confined with metals in metal ores e.g. pyrite, FeS
2
and galena PbS. An ore is a
rock or earth from which metals are extracted.
Physical properties of sulfur:
It is brittle.
It is yellow in color.
It is made up of crown-shaped molecules, each with eight atoms.
It exists in different allotropes.
Chemical properties:
It burns easily with a blue flame in the presence of air to give sulfur dioxide.
)()()(
22
gSOgOsS
It combines with metals to form metallic sulfides e.g.:
.)()( FeSsSsFe
When heated with sulfuric acid it gets oxidized to SO
2
according to the following:
.2242
32)(2)( SOOHaqSOHsS
It combines with non-metals such as H
2
to produce hydrogen sulfide characterized by
the odor of rotten eggs. i.e.
).()()(
22
gSHsSgH
Allotropy.
It is the existence of an element in more than one crystalline form.
Allotropes.
These are different physical forms of an element.
E.g.; the allotropes of carbon are diamond which does not conduct electricity and
graphite which is used to make pencils and it conducts partially.
Allotropes of sulfur include rhombic sulfur, monoclinic sulfur, colloidal sulfur and plastic
sulfur.
40
The major allotropes of sulfur:
Rhombic sulfur and Monoclinic sulfur
a) Rhombic sulfur
o It is also called alpha sulfur (
- sulfur)
o It is a relatively large, yellow translucent octahedral crystal.
o It forms at room temperature hence it is stable at any temperature below
96°C.
o It melts at 114°C and a density of 2g/cm
3
.
b) Monoclinic sulfur.
o It is also called prismatic sulfur or beta sulfur (
-sulfur)
o It consists of needle-shaped transparent crystals.
o It melts at 119°C.
o It has a density of 1.98g/cm
3
.
o It is stable at a temperature above 96°C. It converts to rhombic sulfur below
96°C.
The figure below shows the pictorial difference between Beta and alpha sulfur.
USES OF SULFUR
1. Used in making of sulfuric acid.
2. Used in vulcanization of rubber. Vulcanization is the process of making rubber tough
and more elastic by addition of sulfur. Tyres are made in this way.
3. Used to make special concrete called sulfur concrete which is not affected by acids.
4. It is vital in the production of important chemicals such as carbon disulfide and calcium
hydrogen sulfate which is used in paper industry.
5. Raw material in the manufacture of various drugs and ointments.
6. Used to produce pesticides and fungicides.
7. Used in sterilization (Killing of germs).
8. Used in the manufacture of matches and gun powder. The match head contains, among
other materials, antimony sulfide, SbS. Gun powder is a combination of sodium (v)
41
nitrate, charcoal and sulfur. Fireworks locally known as ‘makombola’ result from this
process.
SULPHURIC ACID, H
2
SO
4
o It is a colorless, odorless oily liquid which has a density of 1.86 g/cm
3
and boils at 338°C.
o It is made by the contact process. This involves series of stages. Firstly, Sulfur is allowed
to burn in air to produce sulfur dioxide. Later, sulfur dioxide produced is exposed to air
under controlled conditions. This process leads to production of sulfur trioxide which is
dissolved in water in order to produce the sulfuric acid. The following are word
equations for the series of reactions in the contact process.
1.
22
SOOS
2.
322
2SOOSO
3.
4232
SOHSOOH
o It is the cheapest acid.
USES OF SUPHURIC ACID:
1) It is used in the manufacture of fertilizers like calcium sulphate, CaSO
4
2) It is used in the extraction of metals and in the pickling process. (Pickling is cleaning
metals with acid prior to painting.)
3) It is a raw material in the manufacture of fibers and plastics.
4) It used in the manufacture of synthetic paints, dyes, and explosives, soaps, detergents
etc.
5) Used as an electrolyte in car batteries.
6) Used in the refining of petroleum.
7) Used in the manufacture of other chemicals.
8) Used in tanning of leather.
9) Used as a dehydrating agent. I.e. it can remove water from a substance. For instance, it
turns blue copper (II) sulphate crystals into white powder by removing water of
crystallization.
4224
5.
42
CuSOOHOHCuSO
SOH
.
It also dehydrates sugar, paper and wood. When sugar is dehydrated, black staff, which
is carbon, remains in the container. The water escapes away. Thus,
OHCOHC
SOH
2112212
1112
42
10) Dilute sulfuric acid reacts with metals to give hydrogen and salts called sulphates.
e.g.
)()(
2442
gHMgSOaqSOHMg
SULPHATES:
o They are ionic compounds.
42
o Examples are Lead sulphate, PbSO
4
, calciumsulphate, CaSO
4
, barium sulphate, BaSO
4
etc.
o Most sulphates are soluble in water.
Tests for sulphates:
o Solubility in water. All sulphates are soluble except those made of Pb, Ca and Ba.
o Add a few drops of dilute HCl and a soluble barium salt like barium chloride, BaCl
2
. A white
precipitate of barium sulphate is produced according to the following equation:
)()(
4
2
4
2
sBaSOSOaqBa
.
USES OF SULPHATES:
I. In fertilizer making. e.g. CaSO
4
for making ammonium sulphate,
424
)( SONH
II. In the manufacture of medicine e.g. MgSO
4
used as laxative Epson salts.
III. In the manufacture of plaster of Paris, POP. E.g. calcium sulphate.
IV. Used in glass and paper making. (Sodium sulphate).
V. Used in making of eye lotion and mouth washes e.g. Zinc sulphate. This is also used in dying of
clothes and healing of wounds.
VI. Other uses include making pigments, purification of metals, electroplating and in diagnostic
medical x-rays.
MANEB QUESTIONS IN THE RECENT THREE YEARS
2015 No 6
a. figure below shows an electron dot and cross diagram of ammonia
i. Name the type of bonding that holds the atoms together. Answer: Covalent bonding
ii. Give a reason for the answer in 6a(i)
Both Nitrogen and hydrogen are non-metals and are sharing electrons
iii. Write the chemical formula for ammonia Answer:
3
NH
B . (i) Mention any three properties of metals
Conduct electricity, Are malleable, Are ductile, Are hard etc
43
(ii) Explain how metallic bonding occurs
Answer: metals have free mobile electrons. There is force of attraction between the sea of free
elections and the positively charged nuclei. This is metallic bonding
8b. In terms of electrical conductivity, explain the difference between “polar” and “non-polar”
covalent molecules.
Answer:
Polar covalent molecules have one end slightly positively charged and the other negatively charged
while the non-polar covalent molecules have neutral ends. This allows polar covalent molecules to
slightly conduct electricity while the non-polar covalent molecules do not conduct electricity.
2014
a. State the three Subatomic particles of an atom
Ans: electrons, protons, neutrons
b. An atom with a mass number of 23 has 13 neutrons. Work out the electron
configuration for the atom
Answer:
ut
c. Table below shows the electrical conductivity of solids A, B, C, D and E when dissolved in water
Compound
Conductivity
A
Does not conduct
B
Conducts
C
Does not conduct
D
Conducts
E
Does not conduct
(i) Classify the compounds as ionic and molecular
Answer: Ionic B and D Molecular: A, C , E
(ii) Give a reason for your answer above
Answer: Ionic compounds form ions when dissolved in water hence they
conduct electricity.
2013
a. Mention any one difference between ‘polar’ and ‘non-polar’ molecules
Answer: Polar molecules have one end slightly positively charged and the other end
slightly negatively charged while non-polar molecules have neutral ends
b. Table below shows atomic numbers, melting points and boiling points of group 7
elements
Number of electrons=mass number-number of neutrons = 23- 13 = 10
Electron configuration = 2, 8
44
Elements
Atomic
Number
Melting point
(°C)
Boiling point
(°C)
Fluorine
9
-220
-188
Chlorine
17
-101
-34
Bromine
35
-7
59
Iodine
53
114
184
(i) Which elements are gases at room temperature?
Answer: Fluorine and Chlorine
(ii) Draw the atomic structure of chlorine (Cl)
Answer
(iii) Why does iodine have a higher melting point than fluorine?
Answer: it is because iodine has a higher atomic number which leads to high
IMF and also iodine is solid at room temperature which means its molecules
are more compacted together than in fluorine hence more heat is needed to
cause the boiling.
(iv) Calculate the number of neutrons in an iodine atom if its atomic mass is 127
2012 6
a. The table below shows electron configurations of element R, S, T, U and V
Element
Electron configuration
R
2,7
S
2,8,6
T
2,8,2
U
2,4
V
2
(i) Which elements in the table belong to period 2 of the periodic table? R and U
Number of neutrons = A – Z = 127-53 = 74
45
(ii) Give a reason for your answer above. Have two shells, a number equal to period
number
(iii) Give a pair of elements that would form an ionic compound when they react. T and U
0r T and S or T and R
(iv) Draw an electron dot and cross diagram for the compound formed when S combines
with U
b. State any three physical properties of halogens
They are colored and they darken going down the group.
They exist as diatomic molecules because two atoms can gain full shells by sharing
electrons with each other.
Their boiling points and melting points increase going down the group because of the
increase in the force of attraction between the molecules. etc
c. Explain what happens if chlorine is mixed with potassium bromide solution.
Displacement reaction would occur. There will be formation of potassium chloride
since chlorine is more reactive that bromine.
2011. 4a
I. State three ways in which atoms attain stability.
By loosing electrons, By gaining electrons, By sharing electrons.
II. Explain how ionic bonding occurs.
Ans: It occurs between a metal and a non-metal. The metal loses electrons to become a cation
while the non-metal accepts electrons to become an anion.
b. Table below shows atomic numbers and boiling points of some elements represented by letters D, Q,
T, X and Z.
ELLEMENT
ATOMIC NUMBER
BOIING POINT (°C)
D
3
1342
Q
13
2467
T
16
445
X
18
-187
Z
19
760
I. Identify any two letters that represent elements which belong to period 3 in the periodic table.
Ans: Q, T or X.
II. Which element is in gaseous state at room temperature (25°C)?
Ans: X
III. What type of bonding would exist when element Q reacts with element T?
Ans: Ionic bond.
IV. Write down the chemical equation for the reaction that would occur between D and T.
2
DTTD
____________________________________________________________________________________
46
3.CHEMICAL REACTIONs1.
OBJECTIVES:
By the end of this topic you should be able to:
Express a chemical reaction by a balanced equation.
Use the mole in relation with other units.
Relate the mass to unit of measurement.
Calculate the empirical formula and molecular formula.
Express concentration.
Prepare standard solutions.
Describe titration and its use.
State molar volume.
Calculate the reacting masses and gas volumes.
Classify reactions as exothermic and endothermic.
Draw energy level diagrams.
Chemical reaction:
It is the rearrangement of atoms to form new species.
Examples of chemical reactions are: rusting of iron, burning, precipitation of solids,
decomposition of substances etc.
Generally chemical reactions are affected by the following factors: concentration,
temperature, particle size and catalysts.
Chemical equation:
It is a quantitative statement used to summarize a chemical change.
Substances found on the left of a chemical equation are called reactants (reagents) and
those on the right are called products.
Types(Examples) of chemical reactions:
Combination or synthesis. E.g. photosynthesis. Decomposition.Precipitation.Combustion
or Burning. Redox (Reduction and Oxidation). Displacement. Neutralization. Nuclear.
Electrolysis. Reversible. etc.
Balancing Chemical Equations:
Rules:
o Identify all reactants and products and write their correct chemical formula on both side
of the equation.
o Indicate the physical states of the reactants and the products: (s) for solid, (l) for liquid,
(g) for gas and (aq) for aqueous.
o Balance the equation by trying suitable coefficients. Don’t change subscripts as this may
change chemical formula.
47
o Balance first, elements that appear least on either side of the equation. Those that
appear in several places should be balanced at the end.
o Check your balanced equation to be sure that you have the same number of each type
of atoms on both side of the equation.
Example:
Balance the following equation:
HBrBrH
22
Answer:
)(2)()(
22
lHBrlBrgH
Exercise:
Balance the following chemical equation.
1.
OHOH
222
2.
22222
OOHOOH
3.
428
SOOS
4.
OHCOOCH
2224
5.
23
OKClKClO
6.
42222
SOHHBrOHSOBr
Chemical Formula:
o It is a formula that expresses composition of a compound in terms of symbols of the
atoms of the elements involved.
Relative Atomic Mass, A
r
o This is given by
A
r
= (Mass of one atom of an element)/(
12
1
x the mass of one atom of carbon).
o Atomic masses are found in the periodic table.
Relative Molecular Mass, M
r
.
o It is given by
M
r
= (Mass of one molecule of a compound)/(
12
1
x the mass of one atom of carbon).
o It is also loosely referred to as molar mass.
Examples: work out the Relative Formula masses or molar masses of the following: (RAM: N =
14, O = 16, Cu = 64, H = 1, Al = 27, S = 32 Na = 23, C= 12).
1. H
2
O 2. Al
2
(SO
4
)
3
3.
NaCl 4. SO
3
5. C
6
H
5
OH 6. Na
2
CO
3
.10 H
2
O
7. CuSO
4
.
Solutions:
1. ( 1x2 + 1x16)= 18 2. (27x2 + 32x3+16x 7) = 262 3. 58 4. 60 5. 54
6. 266 7. 160
Calculating percent composition:
Write down the formula of the compound.
Using a list of RAMs work out its molecular mass.
Write the mass of the element you want as a fraction of the total.
48
Multiply the fraction by 100% to give percentage.
e.g:
Calculate the percent composition of Carbon in Urea, (NH
2
)
2
CO. (RAM: H =1, C =12, O = 16,
N=14).
Ans. Formula is (NH
2
)
2
CO.
Molar mass = 2 x14 + 1x2x2+ 12+16 = 60g/mol
Percent Composition of Carbon = (
%100
60
12
x
) = 20%
THE MOLE CONCEPT:
A mole (mol) refers to number of particles equivalent to 6.02x 10
23
particles.
The mole of a substance is obtained by weighing out the RAM formula mass in grams.
The number 6.022 x 10
23
is called the Avogadro’s number or Avogadro’s constant.
A mole is just like the unit dozen (12), unit (10) or gross (144)
Molar Mass:
This is the mass of one mole of a compound or an element. E.g. magnesium has a molar
mass of 24grams/mole.
If m is a mass of a substance with molar mass M, then the amount of substance (moles),
n is given by
M
m
Molarmass
mass
n
I.e.
M
m
n
where n is the number of moles, m is the mass and M is the molar
mass.
Examples:
a. How many moles are there in
1. 0.25g of CO
2
2. 2g 0f CH
4
3. 40g of NaOH.
Solution:
1.
moles
molegx
g
M
m
n 00568.0
/)21612(
25.0
2.
moles
molegx
g
M
m
n 125.0
/)4112(
2
3.
mole
moleg
g
M
m
n 1
/)11623(
40
b. Find the mass of
1. 0.50 moles of H
2
O
2. 5 moles of CO
2
Solution:
49
1. Molar mass for water is 18 g/mole.
gxnMm 91850.0
2. Molar mass is 44 g/mole hence
gxnMm 220445
STOICHIOMETRY:
This is the analysis of the relationship between the amounts of reactants and products of a
chemical reaction. The coefficients of a balanced chemical equation are called stoichiometric
amounts.
Refer to the following balanced equation for the reaction of nitrogen gas and hydrogen gas to
yield ammonia
)(2)(3)(
322
gNHgHgN
The equation above can be described in different ways like
1. 1 molecule of N
2
gas reacts with 3 molecules of H
2
gas to yield 2 molecules of NH
3
gas.
2. 1 mole of N
2
gas reacts with 3 moles of H
2
gas to yield 2 moles of NH
3
gas.
Since m = nM then in 1 mole of N
2
gas: m = nM = 1 mole x (14 x2) g/mole = 28g.
In 3 moles of H
2
m =nM= 3 moles x (1 x2) g/mole = 6 g
In 2 mole of NH
3
m = nM= 2 moles x (14 x 1+ 1x3) g/mole = 34 g.
3. 28g of N
2
gas reacts with 6g of H
2
gas to yield 34 of NH
3
gas.
In the equation above,
a. How many moles of hydrogen gas will react with 0.5 moles of nitrogen?
b. How many moles of ammonia will be formed from 0.5 moles of nitrogen?
c. How many grams of ammonia can be formed from 42 grams of nitrogen and how
many grams of Hydrogen are required to completely convert that amounts?
Solution:
a. 1 mole of Nitrogen reacts with 3 moles of hydrogen
0.5 moles will react with less mores of hydrogen
molesmolesxn 5.13
1
5.0
b. 1 mole of Nitrogen forms 2 moles of ammonia
0.5 moles of nitrogen will form less amount of ammonia
molemolesxn 12
1
5.0
c. 28 g of Nitrogen forms 34 g of ammonia
42 g will form more
ggxm 5134
28
42
28g of Nitrogen combines with 6g of hydrogen
42g wil combine with more hydrogen
ggxm 96
28
42
EMPIRICAL FORMULA
o It is defined as the formula which shows simplest ratio of atoms of elements.
50
o It tells us which elements are present and the simplest whole number ratio of the atoms
but not necessary the actual number of their atoms present in the molecule.
o For example, the molecular formula of hexene is C
6
H
12
. Its empirical formula is CH
2
.
o Molecular formula is a simple multiple of the empirical formula.
Finding empirical formula, from mass:
Start with the numbers of grams that combine.
Change the grams to the moles of atoms.
From this, one can tell how many atoms combine.
You then write down the formula.
Example 1:
An analysis of an organic compound showed that 5.4g of carbon combined with 0.9g hydrogen
and 0.8g of oxygen. What is the empirical formula of this compound? (RAM: C=12, H=1 and
O=16).
Solution:
Number of moles for C= (m/M) = (5.4/12) = 0.45moles.
Number of moles for H= (m/M) = (0.9/1) = 0.9moles.
Number of moles for O= (m/M) = (0.8/16) = 0.05moles.
Divide the number of moles by the smallest value which is 0.05
.9
05.0
45.0
molesC
.18
05.0
9.0
molesH
.1
05.0
05.0
moleO
Thus the ratio of atoms is 9:18:1
The empirical formula is C
9
H
18
O
Example 2:
Given the following information obtained from an experiment: 79.2% Carbon, 20.8%Hydrogen.
Find the
a. Empirical formula of the compound.
b. Molecular formula given that the molar mass is 80 g/mole.
Solution:
a. Assume that the percentages are masses in grams.
Number of moles for
.6.6
/12
2.79
moles
moleg
g
C
Number of moles for
.8.20
/1
8.20
moles
moleg
g
H
Divide by the smallest value which is 6.6
.1
6.6
6.6
mole
moles
C
molesmoles
moles
H 315.3
6.6
8.20
This means that the ration of Carbon to Hydrogen is 1:3 and the Empirical
Formula becomes CH
3
.
51
b. Find the molar mass for the empirical formula which is 12+1x3= 15 g/mole. Divide the
molar mass of the compound by the molar mass of the empirical formula to obtain the
multiplying factor, x. i.e.
533.5
/15
/80
moleg
moleg
x
The molecular formula = (CH
3
)
x
= (CH
3
)
5
= C
5
H
15
.
Example 3:
Given that Hydro Carbon A contains 85.7 % Carbon. The molar mass of A is 42 g/mole
Calculate the empirical formula of the substance. (RAM C =12, H = 1)
ATOM
MASS
Number of moles
Carbon
85.7g
molesn 1417.7
12
7.85
Hydrogen
14.3g
olesn 3.14
1
3.14
C :
mole
moles
n 1
1417.7
1417.7
H :
moles
moles
n 2
1417.7
3.14
Therefore, the empirical formula is
2
CH
The molar mass from the empirical formula= (12 x 1 + 1 x1) g/mole = 14 g/mole.
The multiplying factor
3
/14
/42
moleg
moleg
x
the molecular formula is (CH
2
)
3
= C
3
H
6
.
CONCENTRATION/ MOLARITY
Concentration is the amount of solute present per unit solvent.
A solute could be a liquid or a solid and the solvent is assumed to be a liquid.
It is given in g/cm
3
, g/liter etc.
E.g. If 50g of Sodium Chloride, NaCl is dissolved in 100cm
3
, what is the concentration of
the solution prepared?
3
3
/5.0
100
50
cmg
cm
g
V
m
C
Molarity
It is the number of moles of solute per cubic decimeters.
Thus, it is the concentration expressed in moles per dm
3
.
Molarity = (number of moles of solute)/ (liters of solution).
Example:
E.g. If 60g of Sodium Chloride, NaCl is dissolved in 100cm
3
, what is the molarity of the
solution prepared?
Need to find the number of moles.
moles
moleg
g
M
m
n 03.1
/)3523(
60
52
Change the volume (100cm
3
) to dm
3
(liters). Do this by dividing the given volume by 1000.
33
1.0
1000
100
dmdmV
.
Molarilty=
Mdmmoles
dm
moles
dmvolume
moles
3.10/3.10
1.0
03.1
,
#
3
33
Molar Solution:
This is the solution that contains 1 mole of a solute per cubic decimeter.
Standard solution:
This is the solution whose concentration is accurately known.
Standard solutions are used in titration and volumetric analysis and in solution
preparation.
DILUTIONS OF SOLUTIONS:
This is the process of preparing a less concentrated solution from a more concentrated one
(usually called stock solution).
Adding more solvent to a given amount of the stock solution changes (decreases)
concentration without changing the number of moles of solute present in the solution. I.e.
number of moles of solute before dilution = number of moles of solute after dilution.
Since molarity is defined as moles of solute/liter of solution the number of moles is therefore
given by
Moles of solute = Molarity x Volume of solution in liters.
Considering the fact that all the solute comes from the original stock solution, one can easily
conclude that C
initual
x V
initial
= C
final
x V
final
.
i.e.
ffii
VCVC
where C
i
is initial concentration, V
i
is initial volume, C
f
is the final
concentration and V
f
is the final (new) volume. The other version of the dilution equation is
ffii
VMVM
where the M stands for molarity and the subscripts simply describe the
molarity as either initial or final.
The bottom line is that number of moles before dilution is equal to number of moles after
dilution.
Examples:
1. 100cm
3
of a solution with concentration of 50g/cm
3
was diluted with water until its final
volume was 250cm
3
. Calculate the new concentration of the solution.
Solution:
V
i
= 100cm
3
, C
i
=50g/cm
3
, V
f
= 250cm
3
, C
f
is the unknown. Using the following equation,
333
250100/50 cmxCcmxcmgVCVC
fffii
33
/20/
250
10050
cmgcmg
x
C
f
53
2. Describe how you would prepare 500ml of 0.8 M HClsolution, starting with a 2.0M stock
solution of HCl.
Solution:
Firstly, calculate the volume of the stock solution which has to be used.
V
i
= ?, C
i
=2M, V
f
= 500ml, C
f
= is the unknown. Using the following equation,
mlMxMxVVCVC
iffii
5008.00.2
mlml
x
V
i
200
2
5008.0
Next step is to dilute 200ml of the stock solution using distilled water (pure or
de-ionized water) to give a final volume of 500ml. Use a 500ml volumetric flask
to obtain accurate measurements.
PREPARING SOLUTION FROM CRYSTALS (SOLID PARTICLES):
Steps:
1. Determine the number of moles contained in the solution by using n = c v
2. Convert the number of moles to mass by using m = n M.
3. Highlight the procedure and materials to be used.
Example: Describe how you can prepare 250 cm
3
of a 1.8 M solution of Sodium hydroxide,
NaOH using sodium hydroxide pellets. (RAM Na = 23, O= 16 and H = 1).
Solution:
Firstly, determine how many moles are contained in the solution. This is done by
molesdmxdmmolesCVn 45.0)1000/250()/8.1(
33
Next, convert the number of moles to mass. Molar mass of NaOH = (23 x1 +16 x 1+ 1x1)
g/mole= 40 g/mole. It follows that
ggxnMm 18)4045.0(
Weigh 18 grams of the sodium hydroxide pellets using a triple beam balance. Using measuring
cylinder, measure 250 cm
3
of pure water and dissolve the sodium hydroxide pellets in the pure
water. Stir or shake properly till all the pellets dissolve.
Molar Volume:
This is the volume occupied by 1 mole of a gas at standard temperature and
pressure.
For reaction of gases, it is more usual to consider the volumes of reactants and
products rather than their masses.
One mole of a molecule of all gasses at standard temperature and pressure, STP
(0°C & 1atm) occupies approximately 22.4dm
3
(22.4 l or 22400cm
3
). This is called
Molar Volume.
At room temperature, 25°C and pressure 1atm, molar volume is 24dm
3
.
Example:
Calculate the volume occupied by 5g of oxygen gas at STP. (RAM O =16).
Solution:
54
One has to find the number of moles of oxygen contained in the 5g.
.156.0
/32
5
/)216(
5
moles
moleg
g
molegx
g
M
m
n
Now 1mole of any gas, at STP, occupies 22.4dm
3
.
Hence volume of gas = 0.156 x 22.4dm
3
= 3.5dm
3
.
TITRATION:
This is a gradual addition of a solution into another until the reaction is complete.
There are different types of titration but the commonly carried out is acid-base
titration.
Acid-Base Titration:
o In an acid-base titration the point at which the acid is completely reacted with or
neutralized by the base is called the end-point or equivalence point.
o End point is signaled by a sharp color change of indicator added to an acid-base mixture.
Indicators are substances that have distinct colors in acidic or basic media.
o An example of an indicator is phenolphthalein which is colorless in acids and pink in
basic solutions.
o The apparatus used in the titration experiments include: conical flask, pipette, burette,
funnels, measuring cylinder, pipette filler, indicators and sometimes a white towel.
o The following is the titration equation used in calculating either volumes or
concentrations of solutions.
b
bb
a
aa
n
xVC
n
xVC
Where
notation
Meaning
C
a
V
a
n
a
C
b
V
b
n
b
Concentration of the acid.
Volume of the acid
Number of moles of the acid
Concentration of the base
Volume of the base
Number of moles of the base
55
Note: the number of moles is obtained from a balanced chemical equation summarizing
the reaction. It is the coefficients of a balanced equation of usually acid and base. This is
called Stoichiometry (the relationship between the amounts of reactants and the products
in a chemical reaction).
Example 1:
o Q: 10cm
3
of hydrochloric acid, HCl neutralized 15 cm
3
of a 0.2M sodium hydroxide (
NaOH) solution.
1) Identify a standard solution and give a reason for your choice.
2) Calculate the concentration of the hydrochloric acid.
o A: 1) it is the sodium hydroxide solution because its concentration is known.
Equation:
OHNaClNaOHHCl
2
. It is balanced.
V
a
= 10cm
3
, C
a
=?, n
a
= 1, V
b
= 15cm
3
, C
b
= 0.2M, n
b
= 1. Using the following:
mole
cmMx
mole
cmxC
a
1
152.0
1
10
3
3
MM
x
C
a
3.0
10
152.0
Example 2:
o Q: In a certain process, 20cm
3
of a 1.2M solution of sodium hydroxide, NaOH was
titrated against sulphuric acid. the process was done three times and the results are
tabulated as below:
# of run
Volume of the H
2
SO
4
.
Volume used, cm
3
Initial volume, cm
3
Final volume,cm
3
1
50
35.1
2
35.1
20.1
3
20.1
5.3
1) Calculate the volume of the sulfuric acid used in each run hence find the
average volume of the titer (the acid).
2) Write a balanced chemical equation.
3) Calculate the concentration of the sulfuric acid.
o A:
1) This is done by filling-in the table as is below:
# of run
Volume of the H
2
SO
4
.
Volume used, cm
3
Initial volume, cm
3
Final volume,cm
3
1
50
35.1
14.9
2
35.1
20.1
15
3
20.1
5.3
14.8
Average volume =
33
9.14
3
7.44
3
8.14159.14
cmcm
56
2)
)(2)()(2)(
24242
lOHsSONaaqNaOHaqSOH
3) V
a
= 14.9cm
3
, Ca=? n
a
= 1, V
b
= 20cm
3
, C
b
= 1.2M, n
b
=2. Using the
following:
moles
cmMx
mole
cmxC
a
2
202.1
1
9.14
3
3
MM
x
x
C
a
81.0
9.142
202.1
The basic procedure is as follows:
You are provided with the following; burette, clamp and clamp stand, measuring cylinder
conical flask, phenolphthalein (or any) indicator, basic solution (egNaOH) and acid,
(egHCl)Theconcentration of either the acid or the base is known.
a. Set up the apparatus as shown below
b. Fill the burette to the mark with the acid (e.g.HCl)
c. Record the volume of acid
d. Measure a given volume of the base e.g.NaOH and transfer it into conical flask
e. Add 2 drops of phenolphthalein(or any) indicator into the conical flask
f. Add the acid gradually, in small amounts from the burette into the conical flask. Swirl the conical flask to mix well
g. Stop adding acid when a color change is observed in the flask
h. Record the volume of acid remaining in the burette
i. Subtract the final volume of acid from the initial volume and record
Initial volume of acid = ________________________
Final volume of acid =__________________________
Volume of acid used = __________________________
j. Write a balanced chemical equation for the reaction
k. Calculate the concentration of either the acid or the base
l. Identify the standard solution in this experiment
m. State any two sources of error in the experiment.( These may include; failure to read the volume changes
accurately, addition of acid whencolor change has taken place, malfunctioning apparatus like burette’s tap not
closing (opening ) properly among others.
57
HEATS OF REACTION:
o There are two kinds of energy namely: potential energy and kinetic energy.
o Potential energy is energy due to position of an object.
o Kinetic energy is energy due to movement.
o Heat is a form of kinetic energy. It is associated with motion of molecules.
o The energy of chemical bonds is a form of potential energy emanating from the
position of atoms and molecules with respect to each other.
o Energy is measured in joules.
o The study of heat changes in chemical reaction is called thermo chemistry.
o The energy stored in chemical bonds is called enthalpy and is represented by H.
o Change in enthalpy is denoted by ΔH (delta H).
o The enthalpy of reaction is the difference between the enthalpies of the products
and the reactants. I.e. ΔH = H (products) - H (reactants).
o ΔH is called the heat of reaction.
o A reaction can either be endothermic or exothermic.
Endothermic reaction:
This is a reaction in which energy is absorbed from the surrounding.
In endothermic reactions
ΔH is positive because the products have higher energy than the reactants.
ΔT (change in temperature) is negative since the final temperature is lower than the
initial temperature.
Examples of endothermic reactions are: thermal decomposition, photosynthesis and
dissolving, polymerization of ethene to poly-ethene, reactions that take place during
food cooking, reduction of silver ions to silver during photography.
Consider the reaction of Nitrogen with oxygen to produce nitrogen dioxide.
)(2)()(
22
gNOgOgN
. Its energy level diagram or energy level profile will be as
follows.
Exothermic reaction
This is the reaction that releases heat to the surrounding.
They are much less common than the endothermic ones.
58
In exothermic reactions
ΔH is negative since the products have lower energy value than the products.
ΔT is positive.
Examples include: fermentation, neutralization of acids by bases, ionization of acids in
water, combustion of fuels and respiration in animal bodies.
Consider the reaction of a hydrochloric acid, HCl and a sodium hydroxide solution, NaOH
according to the following reaction.
OHNaClNaOHHCl
2
And ΔH = -57KJ/mole. The energy level diagram will be
as follows:
__________________________________________________________________________
HOW THE TOPIC HAS BEEN FEATURED BY MANEB IN THE RECENT 4 YEARS:
2015 No 5
a. Give two ways of determining strength of an acid
Answers: By conductivity test, by use of universal indicator, by use of pH meter
b. State any two ways of expressing the concentration of a solution
Answers: (i)as mass of solute per unit volume of solvent. (ii) As number of moles per
volume in liters or cubic decimeters
c. In a titration, 25cm
3
of hydrochloric acid (HCl) of unknown concentration was titrated
against 20cm
3
of 2M sodium hydroxide (2M NaOH) to which phenolphthalein was
added.
(i) Name the standard solution in the titration. Answer: Sodium hydroxide (NaOH)
59
(ii) Give a reason for your answer in 5c(i)
Answer: because its concentration is known
2014 Number 3
a. Define ‘mole’
It a measure equivalent to
23
1003.6
particles or it is a measure of particles
equal to Avogadro’s number
b. Magnesium (Mg) react with oxygen (O
2
) according to the following equations:
)()(2)( sgs
MgOOMg
(i) What does “s” stand for in the equation? Answer: solid
(ii) Balance the equation
)()(2)(
22
sgs
MgOOMg
(iii) If 120g of magnesium reacts completely in excess of oxygen. How many
mores of oxygen are used? (RAM: Mg = 24, O =16)
c. Sodium chloride (NaCl) dissociates in water as follows:
1
)()(
)(2)(
4
kJmolHClNaOHNaCl
aqaq
ls
(i) Name the type of reaction basing on enthalpy change.
Answer Endothermic reaction
(ii) Draw an energy level diagram for the process
120g of Mg
m oles
M
m
n 524/120
If 2 moles of Mg needs 1 mole of O
2 ,
5 moles will need more
Number of moles of oxygen =
molesmol 5.21
2
5
60
2013
3
a. What is an “ empirical formula”?
Anwer: It is a formula that expresses atoms in their smallest ratio
b. Calculate the empirical formula of copper oxide (CuO) with chemical composition of 32g
of copper and 8g of oxygen (RAM: O =16 and Cu = 64)
element
mass
Number of atoms
Cu
32g
moles
M
m
n 5.0
64
32
O
8g
moles
M
m
n 5.0
16
8
Dividing by smallest value :Number of moles of Cu = 0.5/0.5 = 1mole and for oxygen =
0.5/0.5 = 1mole
Therefore empirical formula is CuO
c. Fiugre below is diagram showing the set up of an experiment on titration
(i) What is the function of phenolpthalene indictaor in the experiment?
Answer: to show or signal the end point of the reaction
(ii) Name the standard solution in the experiment
Answer: Sodium hydroxide
61
(iii) Give a reason for your answer above
Answer: because its concentartion is known
d. State any two ways of expressing the concentration of a solution
Answer:
Mass of solute per unit volume of solvent
Number of moles of solute per unit volume of solvent in cubic decimeters
2012
a. Figure below shows an energy level diagram for the reaction between magnesium Mg and
oxygen O
2
.
i. Is the reaction exothermic or endothermic? exothermic
ii. Give a reason for your answer above. It is because the change in heat is
negative which means that heat is given to surrounding making the products
have low energy than reactants.
iii. State the meaning of the arrow in the diagram. It shows direction of reaction in
terms of reactants and products
4.CHEMICAL REACTIONS 2
OBJECTIVES:
By the end of the topic learners should be able to:
Define oxidation and reduction.
Define oxidation number.
Identify reducing and oxidizing agents.
Describe a displacement reaction.
Explain the reactivity series.
Explain rusting/corrosion.
Discuss ways of preventing corrosion.
Describe the process of electroplating.
62
Define an acid and a base.
Describe the formation of a hydroxyl and a hydronium ion.
Identify a conjugate acid-base pair.
Determine strength of acids and bases.
______________________________________________________________________________
Definitions.
REDOX REACTION:
It is a reaction in which both reduction and oxidation occur.
OXIDATION.
It is the loss of electrons.
E.g.
2
2 MgeMg
It is shortened as OIL or LEO.
OIL means Oxidation Is Loss and LEO stands for Loss of Electrons, Oxidation.
It is also defined as an increase in oxidation number. Oxidation number/state is the
amount of charge that an atom would have in an element or compound if electrons
were added or removed completely.
E.g.
2
FeFe
.
It is the addition of oxygen to a substance.
eg.
CuOOCu
REDUCTION.
It is the gain of electrons.
E.g
CleCl
. It is shortened as RIG or GER.
RIG means Reduction Is Gain and GER stands for Gain of Electrons, Reduction.
It is the decrease in oxidation number.
02
CuCu
It is the removal of oxygen from a substance.
Oxidation and reduction occur simultaneously. When one substance gets oxidized the other
gets reduced.
REDUCING AGENT (REDUCTANT)
It is a substance that is capable of donating electrons. It is an electron donor.
OXIDISING AGENT (OXIDANT)
It is a substance that is capable of accepting electrons. It is an electron acceptor.
In short, the substance that gets oxidized is a reducing agent while that which gets reduced is
an oxidizing agent. Redox reactions are also called Electron transfer reactions.
RULES FOR ASSIGNING OXIDATION NUMBERS/STATES
1. The oxidation number of an element in an uncombined state is zero. eg. Na
0
,
Cu
0
, H
2
0
, Al
0
etc.
2. In a compound or a polyatomic ion, the oxidation number of
o Group 1 elements is +1
63
o Group 2 element is +2
o Aluminium is +3
o Oxygen is -2
o Silver is +1
o Fluorine is -1
3. For a neutral compound, the sum of oxidation numbers is zero.
4. The oxidation number of a monatomic ion is the charge on it
5. For a polyatomic ion, the sum of oxidation numbers is equal the charge on it.
Examples.
Work out the oxidation number of the given element in each compound given.
1. KCl, Cl.
10)1( xx
2. KClO
3
(potassium chlorate), Cl.
50)2(3)1( yy
3. NO
2
N
40)2(2 xx
4.
4
NH
N
31)1(4 xx
5. KMnO
4
Mn.
70)2(4)1( xx
Pick out the oxidizing and reducing agent in the following reactions.
I.
02
2)()(2)( AgaqCuaqAgsCu
Cu; from 0 to 2; oxidized hence a reducing agent.
Ag; from1 to 0; reduced hence an oxidizing agent.
II.
KClgClsK 2)()(2
2
K; from 0 to +1; oxidized hence a reducing agent.
Cl; from 0 to -1; reduced hence an oxidizing agent.
III.
AgNaNOAgNOsNa
33
)(
Na; from 0 to 1; oxidized hence a reducing agent.
Ag; from +1 to 0; reduced hence an oxidizing agent.
HALF EQUATIONS
It is an equation that shows electrons involved in either oxidation or reduction.
They are written from an overall reaction.
Examples.
1.
02
2)()(2)( AgaqCuaqAgsCu
)(2)(
20
aqCuesCu
Oxidation.
)(22)(
0
sAgeaqAg
Reduction.
2.
)()()()(
2002
aqZnsCusZnaqCu
64
)(2)(
2
sCueaqCu
. Reduction.
)(2)(
20
aqZnesZn
. Oxidation.
3.
0203
2)(3)(3)(2 AlaqMgsMgaqAl
)(36)(3
20
aqMgesMg
)(26)(2
03
sAleaqAl
WRITING OVERALL/NET/FULL/SIMPLIFIED IONIC EQUATIONS
Steps.
o Balance the number of electrons as we do in linear simultaneous equation.
o Add the two equations together.
Examples:
Given the following half-reactions write down overall ionic equations:
I.
)()(
0
aqNaesNa
)(2)(
02
sMgeaqMg
)(22)(2
0
aqNaesNa
)(2)(
02
sMgeaqMg
)()(2)()(2
02
sMgaqNaaqMgsNa
II.
)(2)(
20
aqFeesFe
(Maneb 2009)
)(2)(
02
sSneaqSn
)()()()(
022
sSnaqFeaqSnsFe
DISPLACEMENT REACTION
A displacement reaction is a reaction in which a more reactive element or ion displaces
a less reactive element from a solution.
Metals displace each other from a solution depending on the nature of reaction.
In order to determine which element will displace which one in solution an
activity/reactivity/displacement/redox series is given.
Potassium, K
Sodium, Na
Calcium, Ca
Magnesium, Mg increasing
Aluminum, Al Reactivity
Zinc, Zn
Iron, Fe
Lead, Pb
Hydrogen, H
Copper, Cu
Silver, Ag
65
Gold, Au
Elements higher up in the activity series displace those below by forcing them to accept
electrons.
The more reactive elements act as reducing agents and the less reactive elements as
oxidizing agent. E.g
I.
)()()(
44
sCuMgSOsMgaqCuSO
II.
)()(
2
sPbaqZnCl
No reaction.
The more reactive an element is, the more electropositive it is.
SIMPLE CELL
o It can be made by dipping two metals of different reactivity in an electrolyte such as
dilute sulfuric acid, lemon juice etc.
o The more reactive metal becomes the cathode while the less reactive metal becomes
the anode.
o The greater the difference in reactivity of the metals the higher is the voltage
produced.
Examples.
Below are results of a set of experiments carried out by a pupil. Study them and answer the
questions that follow. ( Maneb 2001)
4
MgSOCu
No reaction.
4
CuSOMg
Fast reaction
4
CuSOZn
Slow reaction
2
ZnClMg
Moderate reaction
4
MgSOZn
No Reaction.
I.Write the balanced equation for the reaction between Magnesium, Mg and Copper Sulfate,
CuSO
4
.
Ans:
).()()()(
44
sCuaqMgSOaqCuSOsMg
II.Name the oxidizing and reducing agents in the reaction you have written.
Oxidizing agent: Copper
Reducing agent: Magnesium.
III.Arrange the elements in order of decreasing reactivity starting with the most reactive.
Mg, Zn, Cu.
IV.Which combination of metals would give a higher voltage, Magnesium/Zinc or
Copper/Magnesium? Give a reason or your answer.
Magnesium/Copper, because the greater the difference in reactivity the higher the
voltage.
CORROSION
It is a process where by an object is eaten up by a chemical process.
Metals undergo corrosion depending on the condition. For example; iron rusts, Zinc
Aluminum tarnish while green patina forms on copper.
66
RUST
It is the powdery brown layer that forms on the surface of iron.
The chemical name of rust is iron (II) oxide while its chemical formula is Fe
2
O
3
.
The conditions necessary for rusting of iron are water and air (oxygen).
During rusting oxygen gets reduced while iron gets oxidized.
EXPERIMENTS TO SHOW THAT BOTH WATER AND AIR ARE NECESSARY CONDITIONS FOR
RUSTING OF IRON:
The water in P is boiled then cooled to remove air.
Wax or paraffin or Rubber bung prevents air from entering into the test-tube.
Calcium chloride in Q absorbs the moisture or dries the air while the rubber bung prevents
moisture from entering into the test tube.
In the test-tubeR, both water and air are present.
Results:
In test tube P, no rusting takes place because there is no air.
In test-tube Q, no rusting takes place because there is no water.
In test tube R, rusting takes place since both air and water are present.
SOME WAYS OF PREVENTING CORROSION (RUSTING):
1. PAINTING
o It prevents water from entering and since both water and air must be present for corrosion
to occur. No corrosion takes place.
2. OILING/GREASING
It prevents water from entering inside.
3. GALVANISING.
It is the covering of iron by a thin layer of zinc metal using the process of electrolysis.
Zinc only tarnishes and does not corrode all the way through. Galvanizing prevents
water and air from entering the iron.
When part of zinc is damaged the iron gets exposed to moist air and rusting begins. But
since zinc is more reactive than iron, it forces the iron to accept electrons there by
reversing the process of rusting.
4. CHROMIUM COATING.
It is the coating of iron with a chromium metal.
5. TIN-PLATING
67
It is the covering of a metal such as iron by thin layer of tin using electrolysis.
Tin only tarnishes and does not get attracted by acids.
6. CEMENT COATING
7. PLASTIC COVERING
8. SACRIFICIAL PROTECTION
This is a process of using a valuable metal to protect iron from rusting. It is largely
done in ships where by magnesium pates are used to protect rusting. These plates are
replaced over when they are completely corroded.
ELECTROLYSIS:
Electrolysis is a process of decomposing an electrolyte by passing an electric current.
An electrolyte is any substance that conducts electric current in aqueous or molten
(fused) form.
Examples of electrolyte include sulfuric acid, hydrochloric acids, sodium chloride
solution etc.
For a substance to conduct electricity it must have free ions (mobile ions).
Cathode is the negative electrode since it is connected to the negative terminal of a
cell or battery.
Cations (positive ions) are attracted to the cathode where reduction takes place.
Anode is the positive electrode since it is connected to the positive side of the cell or
battery.
Anions (negative ions) are attracted to the anode where oxidation takes place.Below
is the apparatus of electrolysis:
SOME USES OF ELECTROLYSIS:
Production of gases e.g.: O
2,
H
2
, Cl
2
etc.
Production of acids like H
2
SO
4
.
Production of bases e.g. Sodium hydroxide NaOH.
Extraction of metals e.g. Alkali metals (Li, Na, K,) Al, Cu etc.
Purification of metals like copper
Electroplating e.g. galvanization. (Rust prevention by use of zinc).
RULES OF DETERMINING OF WHAT GETS DISCHARGED AT THE ELECTRODES.
68
1. AT THE CATHODE:
o The more reactive an element is, the more it likes to remain in the solution. Ions
of a less reactive element are therefore preferentially discharged at the
cathode.
2. AT THE ANODE:
o If the ions of a halogen (like F
-
, Cl
-
) are present in the electrolyte they get
discharged preferentially.
o If no halogen ions are present hydroxyl ions (OH
-
) are discharged instead
resulting into formation of oxygen and water according to half equation below:
)()(24)(4
22
gOlOHeaqOH
.
Examples.
1. Electrolysis of potassium chloride solution (Maneb 1994).
o At the cathode, hydrogen ions , H
+
get preferentially discharged resulting in the
formation of hydrogen gas according to the half equation:
)(2)(2
2
gHeaqH
.
o At the anode, chloride ions, Cl
-
gets discharged preferentially resulting into the
production of chlorine gas according to the half equation below:
)(2)(2
2
gCleaqCl
.
o Overall effect is that the solution becomes basic due to production of potassium
hydroxide.
2. Electrolysis of molten (fused) potassium chloride (Maneb 1998).
69
o At the cathode, potassium ions, K
+
gets discharged preferentially resulting into
production of potassium atoms which coat the cathode according to the half
equation below:
)()( sKelK
.
o At the anode, chloride ions gets discharged there by forming chlorine gas
according to the half equation below:
)(2)(2
2
gClelCl
.
o Overall effect is that the solution decreases until is all used up.
3. Electrolysis of dilute sulfuric acid or acidified water (Maneb 2003 PII).
o At the cathode, hydrogen ions, H
+
gets discharged there by forming hydrogen
gas according to the following equation:
)(2)(2
2
gHeaqH
………………………1
o At the anode, hydroxyl ions, OH
-
get preferentially discharged resulting into the
production of oxygen and water according to the half equation:
)()(24)(4
22
gOlOHeaqOH
…………………2
o The overall effect is that the solution becomes more concentrated due to the
decomposition of water molecules into hydrogen and oxygen gas.
o The overall ionic equation would be as follows: first multiply the equation 1 by 2
to get
)(24)(4
2
gHeaqH
. Then add the two equations 1 and 2.The result is:
)()(2)(24)(44)(4
222
gOlOHgHeaqOHeaqH
. The net is:
)()(2)(2
222
gOgHlOH
.
4. Electrolysis of copper sulphate solution using carbon electrodes.
o At the cathode, copper ions (Cu
2+)
will be discharged preferentially there by
forming copper atoms which coat on the cathode according to the half-equation:
)(2)(
2
sCueaqCu
.
o At the anode hydroxyl ions get discharged preferentially there by producing
oxygen and water according to the half-equation;
)()(24)(4
22
gOlOHeaqOH
o The overall effect is that the blue color of copper sulfate solution becomes pale
due to decrease in copper ions in the solution.
o The solution becomes acidic due to the formation of sulfuric acid.
o The cathode becomes thicker due to copper atom deposits on it.
5. Electrolysis of copper sulphate solution using copper electrodes.
o At the cathode, copper ions (Cu
2+)
will be discharged preferentially there by
forming copper atoms which coat on the cathode according to the half-equation:
)(2)(
2
sCueaqCu
..
70
o The anode dissolves there by releasing copper ions which go to the cathode via
the solution. The half equation is :
)(2)(
2
aqCuesCu
o The resultant effect is that the blue color of copper sulfate solution remains the
same for some time due to continuous supply of copper ions from the dissolving
anode.
o The cathode becomes thicker due to copper atoms on it.
o The anode decrease in size due to the formation of copper ions from copper
atoms.
ELECTROPLATING
o It is the covering of objects with a thin layer of a metal using electrolysis.
CONDITIONS FOR ELECTROPLATING
1. The metal to be electroplated becomes the cathode.
2. The covering metal becomes the anode.
3. The electrolyte should be a soluble salt of the metal that makes up the anode.
Example:
a) Draw a well labeled apparatus that would be used to electroplate an iron nail with
copper chloride as an electrolyte.
b) Explain what happens during the process of electroplating by the nail. Support the
explanation with relevant chemical equations.
Solution:
a)
71
b) At the cathode, (iron nail) copper ions get discharged preferentially there by forming
copper atoms which coat on the nail. The half equation at the cathode is as follows:
)(2)(
2
sCueaqCu
.
c) The anode dissolves there by releasing copper ions into the solution which go to the
cathode when they get reduced to form copper atoms. The half equation is as follows:
)(2)(
2
aqCuesCu
. (2005 Maneb P1 #7)
SOME USES OF ELECTROPLATING:
Corrosion is prevented e.g. tin-plating, galvanization,
Appearance is improved.
Worn out machinery is replenished.
A hard layer is introduced that reduces wear and tear.
PURIFICATION OF METALS:
Conditions of purification of metals are:
1. The pure metal becomes the cathode.
2. The impure metal becomes the anode.
3. The electrolyte should be soluble salt of the metal being purified.
Example: (Maneb 2009 PII #4)
With the aid of a clearly labeled diagram, describe how impure copper could be purified by
electrolysis. The description should include relevant chemical equations.
72
o At the cathode copper ions get discharged preferentially thereby forming copper
atoms which coat on the pure copper. The half equation is as follows:
)(2)(
2
sCueaqCu
.
o The anode dissolves there by releasing copper ions into the solutions which go to
the cathode where they get reduced to form copper atoms. The half equation is:
)(2)(
2
aqCuesCu
.
o As the process progresses, the cathode becomes thicker due to the copper
deposits on it.
o Deposits of impurities are observed just below the anode.
ACIDS AND BASES. (PROTON-TRANSFER REACTIONS.):
A proton is a hydrogen ion, H
+
. This is the ion that is responsible for the acidic properties
of a substance.
An acid is a substance that is capable of donating protons. It is a proton donor. This is
according to Lowry-Bronsted theory. Examples of acids include HCl, H
2
SO
4
, HNO
3
etc.
A base is a substance that is capable of accepting protons. I.e. proton acceptor. E.g. NH
3
,
NaOH, CuO etc.
The acid (according to Arrhenius) is also defined as a substance that ionizes in water to
release protons while a base is a substance that ionizes in water to accept protons.
PROPERTIES OF ACIDS
I. They have a sour taste.
II. They conduct electricity.
III. They have a pH less than 7. pH is the power of hydrogen ion concentration.
IV. They facilitate rusting.
V. They react with metals to produce salt and hydrogen gas. A salt is an ionic compound
whose cation is a metal and the anion is a non-metal except (OH
-
)and oxide, O
2-
.
e.g.
)()()(2
22
gHsZnClZnaqHC
VI. They react with bases to produce salt and water in what is called neutralization process.
73
OHSONaNaOHSOH
24242
VII. They react with carbonates and hydrogen to produce salts, carbon dioxide and water.
E.g.
OHCOCaClHClCaCO
2223
2
PROPERTIES OF BASES
I. They have a pH more than 7.
II. They have a bitter taste.
III. They have a soppy feel.
IV. They facilitate rusting.
V. They conduct electricity.
VI. They neutralize acids.
Examples include, NaOH, KOH, CuO etc.
CONJUGATE BASES
It is the species that is formed after a proton; H
+
has been removed from an acid.
Examples:
ACID
CONJUGATE BASE
HCl
Cl
42
SOH
4
HSO
4
HSO
2
4
SO
4
NH
3
NH
OH
2
OH
CONJUGATE ACID
It is the species that is formed after a proton has been added to a base.
BASE
CONJUGATE ACID
OH
OH
2
OH
2
OH
3
2
NH
3
NH
3
NH
4
NH
3
HCO
32
COH
42
POH
43
POH
HCl
ClH
2
REVERSIBLE REACTIONS
A reversible reaction is a reaction that goes in either direction.
Once enough products are formed they start to react there by reversing the forward
reaction. The symbol for a reversible reaction is
.
74
When the rate of forward reaction equals the rate of backward reaction, an equilibrium
point is reached and it is symbolized by
.
AMPHOTERIC SUBSTANCE
This is a substance that can act as an acid in one reaction and as a base in another
reaction. Water is a classic example.
Water as a base:
OHCOOCHOHaqCOOHCH
3323
)(
.
Water as an acid:
432
)()( NHaqHONHlOH
HYDROXYL(
OH
) AND HYDRONIUM ,
OH
3
A hydroxyl ion is usually formed when water undergoes auto-ionization (self-ionization)
or when it acts as an acid by donating a proton.
It is also an anion.
eg:
HaqHOlOH )()(
2
432
)()( NHaqHONHlOH
.
A hydronium ion is a cation which is usually formed when water acts as a base by
accepting protons.
ClaqOHlOHHCl )()(
32
STRONG AND WEAK ACIDS
A strong acid is the acid that ionizes fully in water there by releasing more protons.
A strong acid is also a strong electrolyte. I.e. it allows large amount of electric current to
pass through. E.g. HCl, H
2
SO
4
, HNO
3
etc.
A weak acid is an acid that ionizes partially in water there by releasing a few protons.
A weak acid is also a weak electrolyte. e.g. H
2
CO
3
, CH
3
COOH, citric acid etc.
STRONG AND WEAK BASES
A strong base is the base that ionizes fully in water there by accepting more protons.
A strong base is also a strong electrolyte e.g. NH
3
, NaOH, KOH e.t.c.
A weak base is a base that ionizes slightly in water.
A weak base is also a weak electrolyte e.g. Amino methane (CH
3
NH
2
), CuO, Zn(OH)
2
STRENGTH OF ACIDS AND BASES
- This can be determined by using any of the following:
1. Conductivity test
2. Universal indicator
3. PH meter.
1. CONDUCTIVITY TEST.
- The two acids / bases being tested must have same:
75
Concentration
Temperature
Volume
Electrode spacing
Electrode depth
Voltage supply
The apparatus below is used:
Acid or Base
-Record the current for each acid or base in turn. The larger the current reading, the
stronger the acid or base
2. UNIVERSAL INDICATOR.
- Pour equal amount of each acid / base into two separate test tubes.
- Add equal drops of universal indicator of the two acids / base and shake gently.
- Observe the color change and match with those that are found on the pH scale in
order to determine which acid or base is stronger than the other.
3. PH METER.
- This is more accurate and quickest way of determining strength of an acid or base.
HOW THE TOPIC HAS BEEN FEATURED IN THE LAST 4 YEARS OF MANEB EXAMINATIONS
(Find some suggested answers in italics).
2015 8a
Figure below shows the reaction between magnesium (Mg) and silver chloride solution (AgCl)
i. Explain the reaction in figure above
A
76
Answer: Magnesium, being more reactive than silver, will donate electrons to silver
hence silver will be displaced instead magnesium chloride will be formed
ii. Write a chemical equation for the reaction
)(2)()(2)(
2
sAgaqMgClaqAgClsMg
5d.(i) Which metal is used to galvanize iron? Answer Zinc
(ii) Explain how a scratched galvanized iron sheet is protected from rusting.
Answer: When Iron gets oxidized (iron atoms turn into iron ions, the Zinc gets oxidized
too. Since Zinc is more reactive than iron it donates electrons to iron making it reduced
(reversing rusting)
2014 No 3
d.
(i) What is a “weak acid”?
This is an acid that ionizes in water partially there by donating a few
protons
(ii) State any one way of determining the strength of an acid
By conductivity test, By use of universal indicator, By use of a pH meter
(iii) The conjugate acid-base pair for the reaction between water molecules
are
OHOH
32
/
and
OHOH /
2
write an equation for the reaction
)(
)(
3)(2)(2
aqOHOHOHOH
aq
ll
2013 No 3
(i) State any two ways of preventing rusting
By painting, Plasticcovering, Sacrificialprotection, Oiling/greasing,
electroplating
(ii) Explain how rusting occurs
Rusting occurs when iron gets oxidized while oxygen gets reduced in the
presence of water. The iron reacts with water and oxygen to form
hydrated iron oxide.
2012 5
b. Define “electroplating” This is covering a metal with a thin layer of another metal
c. Iron (Fe) displaces copper (Cu) from copper sulphate solution (CuSO
4
).
i. Write down a balanced chemical equation for the reaction
)()()()(
44
sCuaqFeSOaqCuSOsFe
ii. What is the reducing agent in the reaction above? Iron, Fe
iii. Give a reason for your answer. Its oxidation has increased from 0 to +2. It
means it has donated 2electrons to copper.
77
d. What is the deference between “oxidation” and “reduction”? Oxidation is the increase
in oxidation number or gain in oxygen atoms or loss of electrons while reduction is
decrease in oxidation number, loss of oxygen atoms or gain of electrons
e. (i) Define “concentration” of a solution. It is the amount of solute per unit volume of
solvent
(ii) The volume of sodium hydroxide solution, NaOH of concentration 20g/l is increased
from 60cm
3
to 600cm
3
by adding distilled water. Calculate the concentration of the new
solution in g/l
5. FORCES AND MOTION
OBJECTIVES:
By the end of this chapter learners should be able to:
Differentiate between a scalar and a vector.
Add vectors.
Resolve vectors components.
Distinguish between distance and displacement.
Differentiate between speed and velocity.
Define acceleration.
Analyze Velocity-Time graph.
State the Newton’s Laws of Motion.
Discuss the application of the Laws of motion.
Describe the relation between the mass and stretching force.
FORCE:
Force is pull or a push.
SI unit of force is Newton, (N).
Force cannot be seen but its effects can be observed. Thus force is capable of:
Increasing speed of an object (acceleration).
Decreasing speed of an object (deceleration).
Cause a change in direction.
Cause a change in shape (deformation).
lglg
x
V
VC
CVCVC /2/
600
6020
2
11
22211
78
Making an object at rest to start moving.
Making a moving object to come to a halt.
Causing heat production.
Causing sound production.
Force is an example of a vector quantity. A vector quantity is the one that has both
magnitude (size) and direction. Other examples of vectors include; displacement,
velocity, acceleration, momentum, etc.
Other quantities are called scalar. A scalar is any quantity with magnitude only but no
direction. Examples are: mass, speed, distance, height, volume etc.
Vector Representation:
A vector mathematically is a line segment.
It is represented by a straight line with direction.
The length of the line represents magnitude while the arrow gives direction.
To specify a vector, the three terms, value (size), unit and direction must be
given.Eg
A 50N B
This is a vector AB whose magnitude is 50N. It is different from BA.
Vector Addition:
Vectors are added geometrically by either parallelogram rule or triangular rule which
ensures that the direction as well as their magnitude is considered.
A resultant vector is the single vector that represents two or more vectors acting at
particular point.
For example a displacement of 30km eastward followed by a displacement of 40km
south wards produces a resultant displacement of 50km in the direction 53° south of
East.
The above result has been obtained by using Pythagoras theorem and trigonometry (for the
angle).
Note that the tail of one vector is joined to the top of the other vectors to ensure that original
vectors point in the same direction round triangle.
PARALLELOGRAM RULE:
Consider two forces P and Q acting at a point O. The resultant will be vector R.
79
R is a diagonal of the parallelogram of which P and Q are two sides. This is the
parallelogram rule.
The parallelogram rule states that if two vectors, OA andOB are represented in
magnitude and direction by sides
OA
and
OB
of a parallelogram OACB thenOC
represents the resultant
Sometimes the calculation of the resultant vector usually requires the use of the cosine rule as
well as the sine rule for a triangle like:
Cosine rule:
CosAbccba .2
222
Sine rule:
SinC
c
SinB
b
SinA
a
Example:
A force of 3N acts at 60° to a force of 5N. Find the magnitude and the direction of their resultant.
B C
3N R
60° θ 120°
80
O 5N A
20222
49120.53253 NCosxxR
NR 7
Applying Sine rule
3712.0
7
120312033
120
0
0
Sine
R
Sin
Sin
Sin
Sin
R
01
8.21)3712.0(
Sin
The resultant therefore is a force of 7N acting at 21.8° to the 5N force.
QUESTIONS:
1. Find the magnitude and the direction of the resultant of each of the following pairs of vectors.
a. 7N at 90° to 24N
b. 60m East and 40 m West.
c. 20N at 60° to 30N
d. 40m/s West plus 40m/s East.
e. 40N at 110
0
to 50N.
f. 40m/s west and 40m/s east.
g. 60N at 150
0
to 30N.
2. Find the resultant of a displacement of 30m due east followed by a displacement of 70m due south?
COMPONENTS OF VECTORS:
-In some circumstances it is useful to split vector up into two parts labeled components of a vector.
-Just as two vectors may be added to form a single resultant, a single vector may be resolved (split)
into two components at right angles to each other.
-Very often two components are taken horizontally and vertically in which the two parts are called
horizontal component of the vector and the vertical component of the vector respectively.
-In figure below force, F is resolved into two components, F
h
is the horizontal component and F
v
I s
the vertical component.
o The magnitude of the two components are given by the following:
FCosF
h
FSinF
v
81
o Example: Calculate the horizontal and vertical components of a force of 30N acting at 30
0
to the
horizontal component.
NNCosF
h
98.253030
0
NNxSinF
h
153030
0
DISTANCE AND DISPLACEMENT:
DISPLACEMENT:
o It is the distance moved in a certain direction. It is a vector.
o The magnitude of displacement is equal to the distance. This is true for linear motion.
o The SI unit of displacement is meters, m.
DISPLACEMENT-TIME GRAPH:
o The slope (gradient) yields velocity.
o The average velocity is found by dividing the total displacement by the total time.
DISTANCE-TIME GRAPH:
o The slope gives speed.
o Average speed is obtained by dividing the total distance by the total time.
o Horizontal line means no motion. I.e. the body is stationary.
Below is the graph of Distance against time.
Calculate
a. Speed in the first 50 seconds
0
50
100
150
200
250
0 50 100 150
Distance(m)
Time(s)
82
Answer: speed= slope
sm
m
xx
yy
s /2
050
0100
12
12
b. How long did the body stop? Answer (100-50)s = 50 seconds
c. The average speed
Average speed = total distance/ total time hence
sm
s
m
s /33.1
150
200
SPEED, VELOCITY &ACCELERATION:
o Speed is defined as the rate of change of distance. (Distance per unit time.)
o Velocity is the rate of change of displacement. I.e. displacement per unit time.
o Velocity is a vector quantity while the speed is a scalar.
o Acceleration is the rate of change of velocity. It is a vector. It is given in
2
/ sm
.
o The acceleration can be positive or negative. The negative acceleration also known as
deceleration or retardation.
o Mathematically acceleration= (final velocity- initial velocity)/time taken.
o Let a, be the acceleration, v, final velocity, u, initial velocity and t, time taken. We have:
t
uv
a
VELOCITY-TIME GRAPH:
o In this part, we use velocity and speed interchangeably. The same case with distance
displacement.
o The slope of this graph gives acceleration.
o Negative slope implies deceleration.
o Horizontal line means constant velocity (constant speed). The particle is not changing in its
speed. Its acceleration is zero.
o The area under the graph gives the distance travelled.
0
20
40
60
0 10 20 30 40
Velocity (m/s)
Time (s)
83
Work out
a. acceleration in the first 10 seconds
Answer: acceleration = slope therefore
2
12
12
/1
010
/10/20
sm
s
smsm
xx
yy
a
b. Total distance traveled.
Answer: Consider area under the graph.
mssmsmssxmssmsmA 65010)/40/20(
2
1
10/2010)/20/10(
2
1
c. Acceleration in the last 20 seconds
2
12
12
/2
2030
/20/40
sm
ss
smsm
xx
yy
a
NEWTON’S LAWS OF MOTION:
These are laws on which science of mechanics is based.
Inertia is defined as the resistance of a body to start moving or to stop moving once it has
started. Or resistance to change in velocity.
A body of large mass requires a large force to change its speed and direction by a noticeable
amount.
Mass is the quantity of matter in an object. Its SI unit is kilogram, Kg.
Weight is the force acting on an object because of its attraction towards the earth. I.e. force of
gravity.
NEWTON’S FIRST LAW OF MOTION:
It states that” Everybody continues in its states of rest or of uniform motion in a straight line
unless it is acted upon by some external unbalanced forces.
The law is another way of saying that all matter has a built in opposition to being moved if it is at
rest or being stopped if it were in motion.
Example 1: A card
A coin
A container
If the card is flicked sharply the coin stays where it is and falls onto the container while the card
flies off.
Example 2:
84
Occupants of a car which stops suddenly. They jerk forward in an attempt to continue moving.
This is why seat belts are vital.
If a moving body speeds up or slows down or simply changes direction then we know that some
unbalanced forces must be acting on the body.
Force always produces acceleration of a body.
MOMENTUM:
It is the product of mass and its velocity.
Thus momentum = mass x velocity.
The SI unit of momentum is the Kg m/s.
When the resultant external force acting on a system is zero the total momentum of the system
remains constant.
External force = rate of change of momentum=
ma
t
uvm
t
mumv
)(
It means F = ma where m is the mass in Kg
u is the initial velocity
v is the final velocity
t is the time
THE NEWTON’S SECOND LAW:
It states that the rate of change of momentum of a body is directly proportional to the external
force acting on the body.
Consider a body of constant mass m being pushed by a constant force, F and its velocity increases
from u to v in time t. By Newton’s second law we have
Change in momentum/time α Force.
F α ma where a is the acceleration due to the applied force.
As an equation it becomes F = kxmxa where k is a constant which has no unit. The SI unit of
force (Newton) is defined in such a way that k = 1 provided the rate of change of momentum is
also expressed in the relevant SI unit (kg m/s). Force is therefore given by
maF
Examples:
1. A body originally at rest attains a velocity of 30m/s in 6 seconds of its motion. Calculate its
acceleration. If the body weighs 60kg what will be the value of the force that brings the
acceleration?
Solution:
u= 0 m/s, v = 30 m/s , m = 60kg, t =6seconds.
2
/5
6
/0/30
sm
s
smsm
t
uv
a
NsmkgxmaF 300/560
2
85
2. A 200kg car changes its speed from 15m/s to 45 m/s in 10 seconds. Work out the force that cause
the car’s acceleration. m = 200kg, u =15m/s, v =45m/s an t = 10 seconds.
2
/3
10
/15/45
sm
s
smsm
t
uv
a
NsmkgxmaF 600/3200
2
NEWTON’S THIRD LAW OF MOTION:
It states that if a body A exerts a force on body B then B exerts equal but opposite force on A. In
other words it states that for every action there is a reaction.
Examples of cases where the third law is demonstrated:
1. Jumping from a canoe; one exerts a force on it and in return the canoe pushes the individual
away.
2. Walking; your feet pushes the ground backward and in return the ground pushes you forward. The
same applies to acceleration of cars due its tyres pushing the ground backward.
3. A balloon rocket; it moves forward as a result of the push exerted on it by the exhaust gases which
the rocket has pushed out.
Conservation of Momentum:
It state that momentum before collision is equal to momentum after collision.
It is a direct consequence of the Newton’s third law of motion.
Examples:
1. A 60 kg body travels at speed of 5m/s and it hits a stationary body of mass 40kg, they coalesce
and move together. Find the common speed.
Solution: mass= 60kg, we shall P to denote momentum.
mvP
. Total momentum before collision is
skgmsmkgxsmkgxmvP /300/040/560
. Momentum after collision is
kgvvkgkgmvP 100)4060(
. Since momentum before collision = momentum after
collision hence
skgmkgv /300100
sm
kg
skgm
v /3
100
/300
2. A 40 kg body moving at speed of 15m/s hits an 80kg body moving at 5m/s. If they coalesce and
move together. Find the common speed.
Solution:
mvP
. Total momentum before collision is
skgmsmkgxsmkgxmvP /1000/580/1540
. Momentum after collision is
kgvvkgkgmvP 120)8040(
. Since momentum before collision = momentum after
collision hence
skgmkgv /1000120
sm
kg
skgm
v /3.8
120
/1000
86
FREE FALL
A body is said to be in free fall if the only force acting on it is the gravitational
force. (Weight). Thus “free fall” is the kind of falling in which weight is the only
force acting on the falling body.
It then has an acceleration downwards relative to the surface of the earth of
about 9.8 m/s
2
usually rounded to 10m/s
2
and is denoted a letter g.
The gravitational force (weight) is given by
mgF
w
where m is the mass and g
is the acceleration due to gravity and is 10 m/s
2
in the case of free fall.
It is -10m/s
2
if it is upward and +10 m/s
2
if the motion is downwards.
Consider a sphere falling from rest through a resistive medium e.g. a viscous fluid
like oil, glycerin etc.
Sphere falling through a resistive medium
The forces that are acting on the sphere are three namely
1. Its Weight (W)
2. The up thrust (U) due to the displaced fluid
3. The viscous drug (Frictional force).
Initially, the downward force W is much greater than the sum of the upward forces, U+F
i.e. W > U+F and the sphere accelerates downwards. As the velocity v increases so too
does the friction force and eventually U+F equals W. i.e. W- (U+F) = 0. The resultant
force is zero.
The sphere continues to move downwards but because there is now no net force acting
on it, its velocity has a constant maximum value known as its terminal velocity.
Terminal velocity is the maximum velocity of a freely falling body in a fluid when the
opposing forces are equal in magnitude but opposite in direction.
Acceleration of a falling object is not uniform, because the resultant force decreases as
the velocity decreases.
87
Terminal velocity
(Speed, m/s)
Time (s)
Heavy objects, whatever their size are only slightly affected by air resistance.
Weight, W and Up thrust, U do not change during the falling. They remain constant.
Friction force is the one that changes.
Friction force depends on the following:
a. Size of the object.
b. Speed of the object in the medium
c. Viscosity of the fluid.
Viscosity:
It is the resistance of fluid to flow or how easily a fluid flows.
Water is less viscous than cooking oil. It means a small metallic sphere will face less
resistance when falling in water than in oil.
FACTORS WHICH AFFECT TERMINAL VELOCITY:
a) Size of the object:
o A small and dense object such as steel ball-bearing has a high terminal velocity.
b) Shape of the object:
o A light object e.g. a rain drop or materials with a large surface area like open
parachute has a low terminal velocity and it only accelerates over a comparatively
short distance before air resistance equals its weight.
c) Viscosity of the fluid:
o The more viscous the fluid is the smaller the terminal velocity. A steel ball bearing
will have a larger terminal velocity in water than in oil
Falling in Vacuum:
In a vacuum, a falling body experiences only gravitational force hence it is under free fall. Its
acceleration is due to gravity whose value is 10m/s
2
. If two or more objects of similar mass are
released at the same time and from the same height then they will reach at the bottom at the
same time. They will all be under the influence of gravitational force (weight).
However, in air, the same objects will reach bottom at different times if their mass is the same
but have different shapes. Shape has an effect on the terminal velocity. Consider the two
diagrams below
88
A feather a coin
Air vacuum
A B
The coin and the feather are released at the same time. In A, they will reach bottom
differently due to different levels of air resistance they experience. In B, the two will reach at
the bottom at the same time since they will be under free-fall.
Falling of Parachute
A parachute achieves two terminal velocities as it falls down the air.
When a skydiver is falling with parachute un-open, he accelerates because of the
downward pull of gravity on him/her
As the speed increases so too does the air resistance
He/she reaches terminal velocity when the opposing forces become equal
With parachute open
The opening of the parachute increases the surface area hence an increase in the
air resistance. This results into slowing down of the parachute
89
As he slows down the air resistance becomes equal to weight, thus second
terminal velocity is reached. This terminal velocity is lower than the previous one
and is safe for landing
FORCE AND MASS
Stretching Force:
When a spring is stretched the difference between its stretched and un-stretched length
is called the extension.
In a spring stretching experiment, the spring is fixed on one end and stretched in stages
by hanging more and more standard masses from the other end.
The extension in each case in measured.
When a graph of extension against stretching force is plotted it is found to be a straight
line passing through the origin and is straight up to a certain point say, E.
It can be concluded that over a certain section of the graph, the extension is directly
proportional to the stretching force. This means that doubling the stretching force the
extension also gets doubled. Mathematically:
eF
keF
eFK /
Where F is the stretching force an e is the extension. K is the proportionality constant or
the force constant.
Elasticity:
This is the ability of a body to regain its original shape after being stretched.
A material is elastic if it is able to regain its original shape after being deformed. e.g. a
spring, rubber band
Elastic limit, E:
This is a point which marks an important point in the behavior of a spring. Beyond E, the
spring becomes permanently stretched.
Hooke’s Law
90
A material is said to obey Hook’s law if its extension is directly proportional to the
stretching force.
Hooke’s law states that the stretching force of a spring is directly proportional to the
extension within the elastic limit.
E
Extension
(mm)
0 Stretching force (N)
Example:
A spring is stretched 10mm by a weight of 2N. Calculate
a. The force constant, k of the spring
b. The weight W of an object which causes an extension of 80mm.
Solution:
a.
mmNmmNeFK /2.010/2/
b.
NmmxmmNkeF 1680)/2.0(
HOW THE TOPIC HAS BEEN FEATURED IN THE LAST 4 YEARS OF MANEB EXAMINATIONS
(Find some suggested answers in italics).
2015No 7
a. State the difference between ‘distance’ and ‘displacement’
Distance is a scalar while displacement is a vector quantity
Rate of change of distance gives speed while rate of change of displacement gives
velocity
b. State any two examples of vector quantities
Answer: force, displacement, velocity, acceleration, momentum etc
c. A 15Kg box is pulled horizontally with force of 90N. Calculate the acceleration of the box
Answer:
2
/6
15
90
1590 sm
Kg
N
aaKgNmaF
2014 4
a. Name two forces that act against the motion of a stone thrown vertically into the air
Answer: Weight and frictional force
b. State two ways of adding vectors
Answer: By triangular rule and by parallelogram rule
c. Figure below is a diagram showing a boat being pulled by two forces A and B at right
angles to each other. A resultant force of 40N is produced at an angle of 60° to force A
91
Calculate the force A
2013
a. Define ‘terminal velocity’
This is the maximum speed reached by a falling body when opposing forces are
equal in magnitude but opposite in direction
b. Mention three factors that affect the terminal velocity of an object falling in a liquid
Viscosity of the liquid, Shape of the object/ surface area, Mass of the object
c. Figure below shows a journey made by a motorist
Find the horizontal component
NA
N
A
2060cos40
40
60cos
92
Calculate the distance covered by the motorist from A to C
2012 7
a.
2012 No 7
a. State the Newton’s third law of motion
It states that if body A exerts force on body B, B will exert equal but opposite force on
body A ( i. e for every action force there is a reaction force)
b. Give two properties of vector quantities. Have magnitude and direction
c. Mention one method of adding vectors acting at an angle to each other. By parallelogram
vector law or by triangular vector rule.
d. A car decelerates at a rate of 3m/s
2
for 5 seconds. If the initial speed is 20m/s, calculate
the final speed.
6.ORGANIC CHEMISTRY 1
OBJECTIVES:
By the end of this chapter learners should be able to:
State the names of the first ten un-branched primary alkanols.
Draw the structure of the first ten primary alkanols.
Write the general and molecular formula of the first ten alkanols.
Describe the production of ethanol by fermentation of sugars
Describe the oxidation of akanols to carboxylic acids.
State the names and draw the structures of the first un-branched carboxylic acids.
Write the general and molecular formula of the first ten carboxylic acids.
Explain why carboxylic acids are classified as acids.
Give examples of natural sources of carboxylic acids.
Describe the reaction of ethanoic acids and ethanol to form an ester.
Use flow diagram in identifying organic compounds.
Apply the general formula in problem solving.
Describe the melting and boiling points in relation to strength of IMF.
Deceleration = - acceleration.
smvv
v
t
uv
ad
t
uv
a /515203
5
20
Distance = Area under graph
msmssAD 5.671595.0/15)63(
2
1
93
Relate the differences in conductivity of alkanols and carboxylic acids to their functional
group
ORGANIC CHEMISTRY:
o This is the branch of chemistry that deals with carbon compounds.
o Carbon forms more compounds than any other element. It forms four covalent bonds.
ORGANIC COMPOUNDS:
o These are compounds with carbon atoms. The carbon in the compound is often bonded
to elements such as hydrogen, oxygen, nitrogen and other elements.
o Hydrocarbons are the organic compounds with carbon and hydrogen atoms only. E.g.
methane CH
4
, ethane C
2
H
4
, nonane C
9
H
20
etc.
HOMOLOGOUS SERIES:
o This is a family of organic compounds which have similar structure, name endings and
properties.
o They show a trend in physical properties and their formula can be represented by a
general formula.
o Most of these they contain a functional group. A functional group is the atom or group
of atoms which are responsible for the characteristic reaction of a compound. This
definition does not apply to alkenes for they have a double bond as a functional group.
o Examples of homologous series (organic families) include:
i. Alkanes C
n
H
2n+2
ii. Alkenes C
n
H
2n
iii. Alkanols C
n
H
2n+1
OH
iv. Carboxylic acids C
n
H
2n+1
COOH
v. Esters R’COOR
NOMENCLATURE (NAMING SYSTEM)
The following prefixes are used
Number of carbon
atoms
1
2
3
4
5
6
7
8
9
10
Prefix
Meth-
Eth-
Prop-
But-
Pent-
Hex-
Hept-
Oct-
Non-
Dec-
ALKANOLS (ALCOHOLS):
o It is a homologous series of organic compounds whose general formula is C
n
H
2n+1
OH.
o They contain the functional group –OH which is called the hydroxyl group.
94
o According to the International Union of Pure and Applied Chemistry ( IUPAC) system,
alcohols are named by taking the name of alkane with the same number of carbon
atoms and changing the ending from –ane to –anol.
o They increase by –CH
2
-
o The table below shows names, formula of the first ten un-branched alkanols:
n
NAME
FORMULA
STRUCTURE
1
Methanol
CH
3
OH
2
Ethanol
C
2
H
5
OH
3
Propanol
C
3
H
7
OH
4
Butanol
C
4
H
9
OH
5
Pentanol
C
5
H
11
OH
6
Hexanol
C
6
H
13
OH
7
Heptanol
C
7
H
15
OH
95
8
Octanol
C
8
H
17
OH
9
Nonanol
C
9
H
19
OH
10
Decanol
C
10
H
21
OH
ETHANOL:
o It is the best known of the alkanols. In fact it is often called alcohol.
o IT is a good solvent; it dissolves many substances that are not soluble in water. It also
evaporates quickly so it is used in glues, paints, varnishes, synthetic drags. Printing inks,
deodorants (perfumes), explosives, dyes, aftershaves etc.
o It is also a raw material for making other substances such as synthetic rubbers and
flavorings.
o It is also well known for making people drunk. Beer, wine and all alcoholic drinks contain
some ethanol.
o Methylated spirit (Meths) is mainly ethanol, mixed with a little methanol which is
poisonous. The coloring is added and a horrible taste is to stop people drinking it.
Ethanol in which other substances have been added is called denatured alcohol.
PROPERTIES OF ETHANOL:
1. Clear colorless liquid that boils at 78°C.
2. It is miscible in water (dissolves in water).
3. Burns in air (react with oxygen in combustion process).
4. React with sodium to form sodium ethoxide and hydrogen gas.
25252
222 HONaHCNaOHHC
5. Can be dehydrated to form ethene.
6. If left standing in air, gets oxidized to form ethanoic acid
COOHCHOHHC
O
352
7. It does not conduct electricity.
96
PREPARATION OF ETHANOL:
1. By fermentation:
o Fermentation is a reaction in which glucose is changed into alcohol and carbon dioxide by
the action of enzymes in yeast.
o The ethanol in alcoholic drinks is made by fermenting carbohydrates in form of glucose
obtained from fruits e.g. (grapes, vegetables, cereals, maize and burley).
o During fermentation, the enzymes called zymase, catalyzes the fermentation of glucose
into ethanol and carbon dioxide. The equation below summarizes the fermentation
process:
2526126
22)( COOHHCaqOHC
zymase
o When beer is made from barley by fermentation, the barley grains are crashed and
soaked in water to extract the glucose. The liquid is strained and yeast of fungus plant is
added. The yeast converts the glucose by ethanol.
o Fermentation is allowed to go on until enough ethanol has formed. The beer is then
carefully heated to kill the yeast.
o Ethanol can also be prepared using local technology. Figure below shows an indigenous
way of preparing ethanol in form of kachasu by distillation process.
o The pure ethanol is obtained by distillation.
2. Hydration of ethene:
o This involves mixing ethene with water (steam) and passed over a catalyst. The
following addition reaction takes place.
OHHCOHgHC
catayst
52242
)(
o The product becomes a solution of ethanol in water. Most of the water is then removed
by fractional distillation. The mixture of water and ethanol is then heated at about 78°C.
Figure above illustrates how the process is done.
GENERAL USES OF ETHANOL:
1. Used as a source of fuel.
2. Used as essential ingredient of alcoholic beverages.
3. Used as a raw material in pharmaceuticals, perfumes and flavorings.
97
4. Used as a solvent for many organic substances.
5. It is an intermediate in the manufacture of other chemicals such as acetaldehydes.
PHYSICAL PROPERTIES OF ALKANOLS:
a. Solubility:
o All the alkanols are solubility in water. The higher the proportion of –OH group
to the alkyl group, the more soluble the alkanols become and vice-versa.
o The lower members; methanol, ethanol, propanol and butanol are completely
soluble in water while higher ones have a less solubility rate.
b. State at room temperature:
o Aliphatic (straight chain) alcohols with less than 12 carbon atoms are liquids at
room temperature.
c. Melting and boiling points:
o They have higher melting and boiling points than those of alkanes of comparable
molecular mass. This is due to the strong polar bond –OH which needs strong
heat energy to be broken. Boiling points and melting points increase with an
increase in molecular mass.
E.g.
C
3
H
8
-42°C
C
2
H
5
OH
78°C
C
8
H
18
126°C
C
7
H
15
OH
180°C
o In the table above, masses of the propane and ethanol are almost within each
other. 44 for propane and 46 for ethanol and yet they have a very big difference
in boiling points. The same applies to octane and heptanol.
d. Good solvents:
o The polar –OH group enables them to dissolves NaOH and KOH as well as non-
polar solutes such as hexane.
e. Non-electrolyte:
o They do not conduct electricity and as such they are non-electrolytes though
they ionize slightly.
CHEMICAL PROPERTIES:
1. They are amphoteric:
They behave as acids and bases due the presence of the –OH group.
2. They react with highly electropositive metals like sodium and potassium to give off
hydrogen gas and alkoxides.
98
25252
222 HONaHCNaOHHC
3. They react with alkanoic acids to produce esters.
4. They undergo oxidation by oxygen and other reducing agents to form alkanoic acids.
USES OF ALKANOLS
1. Used as a source of fuel.
2. Used as essential ingredient of alcoholic beverages.
3. Used as a raw material in pharmaceuticals, perfumes and flavorings.
4. Used as a solvent for many organic substances.
5. It is an intermediate in the manufacture of other chemicals such as acetaldehydes.
6. Used in ester production
ALKANOIC ACIDS (CARBOXYLIC ACIDS):
They consists of a carbon atom bonded by an oxygen atom and by a single bond to a
hydroxyl group which may be represented as
Thus the functional group of carboxylic acids is the carboxyl –COOH.
Their general formula is C
n
H
2n+1
COOH.
According to IUPAC nomenclature of the alkanoic acids, the –e of the corresponding
alkane is replaced with-oic and acid is added as a second word i.e. the –ane is changed
to –anoic acid.
They increase in size by –CH
2
-
They are weak acids.
NAME
FORMULA
STRUCTURE
M.P(°C)
B.P(°C)
Methanoic acid
(Formic acid)
HCOOH
8
100
Ethanoic acid
(Acetic)
CH
3
COOH
17
118
Propanoic acid
C
2
H
5
COOH
-21
140
99
Butanoic acid
C
3
H
7
COOH
-5.5
163
Example 1
Write down the name and formula of an alkanoic acid with 4 carbon atoms
Name: Butanoic acid
Molecular formula
COOHHC
73
Example 2
Write down the name and formula of an alkanoic acid in which n = 4
Name: Pentanoic acid
Molecular formula
COOHHC
94
SOURCES OF CARBOXYLIC ACIDS:
Carboxylic acids are found in both plants and animal kingdom.
Examples of natural sources of carboxylic acids are listed in the table below:
SOURCE
NAME OF THE ACID
Citrus fruits
Citric acids
Sour milk, breast milk, cow’s milk
Lactic acid
vinegar
Ethanoic acid
Sour wine
Ethanoic acid
Rancid butter
Butanoic acid
Gum benzoic
Benzoic acid
Sweet potatoes
Oxalic acid
Ground nuts, bees and beetles
Methanoic acid
Physical properties of Carboxylic acids:
Long chain acids are liquids at room temperature.
Melting and boiling points increase with an increase in molecular mass.
Are soluble in water. However, solubility decreases as number of carbon atoms in the
molecule increases.
Chemical properties of the Carboxylic acids:
They react with carbonates and hydrogen carbonates to form salts and rebate carbon
dioxide.
OHCOCOONaCHCONaCOOHCH
223323
2
They neutralize alkali to form corresponding salts and water.
OHCOOKCHKOHCOOHCH
233
100
Their solutions affect color of indicators.
They react with alkanols to form esters
OHHCOOCCHCOOHCHOHHC
2523352
They undergo chlorination with exception of formic acid. E.g. they react with
phosphorous pent chloride, PCl
5
to yield chloride.
HClPOCCOClCHPClCOOHCH
3353
They undergo decarboxylation reaction (removal of CO
2
) to alkanes. e.g.
OHCONaCHCOONaCHNaOHCOOHCH
heatheat
232433
.
USES OF CARBOXYLIC ACIDS
Manufacture of soaps, Used to make dyes and perfumes eg. Acetic acid, Used as
solvent, Used to preserve food stuffs, Used as food flavors in salads and soft drinks,
Used to make esters
ESTERS (ALKANOATES):
Alkanoic acids react with alcohols to form pleasant smelling compounds called esters.
The reaction in which akanols react with carboxylic acids in the presence of sulfuric acid
to produce esters and water is called “Esterification”.
When naming esters, the alcohol part is named first followed by the name of carboxylic
acid modified to change in –noate.
Examples are given in the table below:
NAME OF ESTER
FORMULA
AKANOL USED
CARBOXYLIC ACID USED
Ethyl-ethanoate
CH
3
COOCH
2
CH
3
Ethanol
Ethanoic acid
Propyl-methanoate
HCOOCH
2
CH
2
CH
3
Propanol
Methanoic acid
Octylethanoate
CH
3
COOC
8
H
17
Octanol
Ethanoic acid
STRUCTURE OF SOME ESTERS:
Propyl-mathanoate Ethyl-ethanoate
Many esters come from plants like pineapples, olive oil, coconut oil, ground nuts etc. All
these have mixtures of natural esters that contain fat acids and glycerol.
Esters are used as solvents for many organic compounds, used in the production of
perfumes, soaps, cosmetics and food flavorings. They are a source of energy.
FLOW DIAGRAMS USED IN IDENTIFYING ORGANIC COMPOUNDS:
Recall from JCE work on properties of alkenes and alkanes
101
ALKANES
General formula is
22 nn
HC
Are called saturated hydrocarbons because all bonds are completely used up
Are called paraffin meaning, less reactive
Are insoluble in water
Do not conduct electricity
The first few members exists as gases at room temperature
Undergo combustion
OHCOOHC
22283
435
Get involved in substitution reaction
Eg methane + Oxygen →mono-chloromethane and hydrogen chloride gas
)(
324
gHClClCHClCH
Eg methane + Oxygen →dichloromethane and hydrogen chloride gas
)(
2224
gHClClCHClCH
Eg methane + Oxygen →tri-chloromethane and hydrogen chloride gas
)(
324
gHClCHClClCH
Eg methane + Oxygen →Tetra-chloromethane (Carbon tetra chloride) and hydrogen
chloride gas
)(
424
gHClCClClCH
ALKENES
General formula is
nn
HC
2
Are called unsaturated hydrocarbons because of carbon- carbon double bond. more
atoms could be added to them when the double bond breaks
Are recognized due to carbon- carbon double bond
Are insoluble in water
Do not conduct electricity
The first few members exists as gases at room temperature
Undergo combustion
OHCOOHC
22263
6692
Get involved in addition reaction
Egethene + bromine→1,2-dibromothethane
242242
BrHCBrHC
102
Get involved in polymerization
Using their chemical properties, one can carry out simple experiments to identify
organic compounds like alkanes, alkenes, alkanols and alkanoic acids since all these are
color-less, they cannot be identified by a mere sight.
a) Solubility test:
o Alkanes and alkenes are insoluble in water. Two layers are observed when they
have been mixed with water.
o Alkanols and alkanoic acids with few carbon atoms are soluble in water. The
solubility decreases as the carbon chain increases,.
o This test involves putting/ addition of 2 to 10 drops of water to two test tubes
having the alkanes and alkenes.
b) Bromine test:
o Add 2 to 10 drops of bromine solution (brown in color) to alkanes and alkenes.
o The mixture of alkenes and bromine is color-less, an indication that addition
reaction takes place. The result is a brown color in the mixture of alkanes and
bromine.
c) Acid test:
o Add about 10 drops of dilute sodium hydroxide, NaOH in two separate test tubes
and add 1 or 2 drops of phenolphthalein indicator. It turns pink.
o Add one of the chemical into one of the test tube. A color-less mixture will
indicate it’s an acid. Pink will signal presence of an alkanol.
d) Combustion test:
i. complete combustion:
It occurs in plenty of oxygen, Produces a lot of energy, Produces blue flame, Less
smoke is produced, Almost soot-less, Produces water and carbon dioxide
ii. Incomplete combustion
It occurs in insufficient supply of oxygen, Produces less amount of energy
Produces yellow flame, more smoke is produced, produces more soot
Produces Carbon and carbon monoxide in addition to water and carbon
dioxide
Note: organic compounds with fewer carbon atoms burn very well than those with larger
number of carbon atoms.
o Figure below shows a flow diagram that can be used to identify which substance is an
alkene, alkane, alkanol or carboxylic acid.
103
For example: Use a flow diagram on how the following can be distinguished from each other:
Ethanol, Ethanoic acid, Ethane and Ethene.
7. ORGANIC CHEMISTRY 2
OBJECTIVES:
By the end of this chapter learners should be able to:
Define isomers.
Draw structure of isomers.
Name the isomers.
Write down the condensed structural formula.
Differentiate monomers from polymers.
Describe the types of polymerization.
Describe the properties of polymers,
State the uses of polymers
Explain different types of plastics.
Discuss waste management.
______________________________________________________________________________
ISOMERISM:
It is the existence of different compounds with the same molecular formula but
different structural formula
104
Isomers are compounds with the same molecular formula but different structural
formula.
The isomers do have different melting and boiling points.
A structural formula gives a clear arrangement and bonding of the atoms. The number
of atoms is also specified.
The names of isomers are given according to the IUPAC system of nomenclature.
The following are guidelines for naming isomers:
- Name the longest un-branched carbon chain.
- Name the substituent group.
- Give the position of the substituent groups.
During numbering of the long chain carbons, start with the carbon that is closer to a
functional group.
Examples:
The following are isomers of pentanol:
Pentan-1-ol Pentan-2-ol Pentan-3-ol
Isomers of butanol , C
4
H
9
OH
Isomers of Pentane, C
5
H
12
.
Normal pentane 2, 2- dimethyl propane 2-methyl butane
Isomers of Hexene C
6
H
12
: just three of them but they are more
105
Carbon rings:
Cyclo-alkanes:
These are alkanes in which carbon atoms are arranged in a ring. Egcyclopropane, C
3
H
6
,
cyclobutane, C
4
H
8
, cyclopentane, C
5
H
10
, cyclohexane, C
6
H
12
.
Cycloalkenes: Examples include benzene C
6
H
6
and cyclohexene,C
6
H
10
.
Cyclo-alkane
Formula
Structure
Cyclo-propane
C
3
H
6
,
cyclobutane
C
4
H
8
CONFORMATIONS:
These are structures which are identical except their shape.
Normally, they arise due to bending and twisting of the carbon-carbon bonds.
Examples:
CONDENSED OR REDUCED STRUCTURAL FORMULA:
It is a formula which indicates the structure of the molecules and how the different
atoms are joined together.
Examples:
106
Write down the condensed formula of the following:
1) C
3
H
8
. Ans: CH
3
CH
2
CH
3
2) C
5
H
10
Ans: CH
3
CH
2
CH
2
CHCH
2
. Or CH
3
CHCHCH
2
CH
3
.
3) C
4
H
9
OH. Ans: CH
3
CH
2
CH
2
CH
2
OH
POLYMERISATION:
This is a process in which small units of molecules combine to form large and complex
molecules.
The long chain molecule formed is called a polymer.
A monomer is a unit molecule which forms a polymer.
In a polymer, hundreds or thousands of unit molecules (monomers) of a given substance
are bonded together to form polymers (macro-molecules).
CLASSFICATION OF POLYMERS:
a. Natural polymers (Bio-polymers):
Examples are: carbohydrates, proteins, natural rubber, deoxyribonucleic acid (DNA) etc.
b. Synthetic (artificial) polymers:
These are man-made polymers which include polythene, nylon, terylene, human hair
poly vinyl chloride (PVC) etc.
TYPES OF POLYMERIZATION:
a. Addition polymerization
b. Condensation polymerization.
a) Addition polymerization:
It is also known as poly-addition.
It involves monomers with carbon-carbon double bond
There is no loss of part of monomer molecules
It involves a successive addition of repeating monomer molecule.
Unsaturated ethene molecule (CH
2
=CH
2)
and its derivatives (Chloroethene/ vinyl
Chloride) undergo self-addition reactions to give high polymers of linear structures.
Ethene which forms polythene Propene which forms Poly-propene
107
Examples of polymers formed in this way are polythene, PVC, polystyrene, Perspex,
polypropene, Teflon etc.
It takes place when the carbon-carbon double bond breaks under the presence of a
suitable catalyst.
b) Condensation polymerization:
This is a polymerization which involves elimination of part of monomer molecules.
It involves different monomer molecules.
It involves monomers with many functional groups
Examples of small molecules that are eliminated from the functional group of
adjacent monomers include water, H
2
O and ammonia, NH
3
.
Examples of polymers formed in this way are Terylene and nylon.
CATEGORIES OF PLASTICS:
a. Thermoplastics.
Are also known as thermo softeningplastics.
108
Examples include polythene and PVC
They are flexible and do not easily break. This is due to the absence of cross-
linking bonds.
Stretch under tension because molecules slide over each other.
Melt at low temperature.
They can be molded into shape after they are already made. (Easy to recycle).
b. Thermo setting plastics.
They are also known as thermo sets.
Examples include melamine and Bakelite.
When heated, they char/break down rather than melt.
They are rigid and break rather than bends. If dropped, the giant structure
breaks at weak points. They have cross-linking bonds which make them strong.
They are molded into shape while they are being made because the shape
cannot be altered anymore.
USES OF POLYMERS:
a. Category of plastics:
I. polythene(polyethene)
o This is made from ethylene (ethane)
o It is tough and flexible
o It is resistant to most chemicals such as acids and bases.
o It is used for making polythene bags, wrappings and covering as well as
water pipes.
o It is also used in the manufacture of flexible bottles, films and insulation
wires.
II. poly vinyl chloride (PVC):
o It is used for electrical equipment, floor tiles, rain coats, shower curtains
and hose pipes.
III. poly propene
o It is durable and used for light weight bottle crates and carpets.
IV. poly tetra fluoroethene (PTFE/Teflon):
o It is resistant to heat and some chemicals.
o It used for electrical insulation, greaseless bearings in liners for pots and
pans.
V. Perspex:
o It is a tough, clear and transparent plastic.
o It is used in aircrafts industry, used instead of glass in the manufacturing
of prisms, lenses, lab equipment, surgical tools and dentures (false teeth).
b. Category of fibers:
I. Terylene:
109
o This is a polyester fiber formed as a product of condensation
polymerization.
o It has a considerable strength, resistant to chemicals and insects such as
months.
o It used for designing of clothes, sheets, tires, nets.
II. Nylon:
o It is a white solid with much greater strength than any natural fiber.
o It is used for clothing, tents, ropes, tires, nets etc.
o Other fibers include viscose (artificial silk), cellophane and Orion.
c. Proteins:
o Are built up by polymerization of amino-acid units
o Examples include keratin, insulin, pepsin, hemoglobin and antibodies.
d. DNA (Deoxyribonucleic acid):
o This is responsible for heredity.
o IT transmits characteristics of a parent cell to a daughter nucleus.
e. Carbohydrates:
o Main source of energy.
o Classified as; monosaccharide, disaccharides and polysaccharides.
WASTE MANAGEMENT (WASTE DISPOSAL):
Wastes, if not handled well, could provide fertile grounds for micro-organism. There are a
number of ways of disposing off of the wastes.
a) Recycling:
It is the use of wastes as raw materials for new production.
Metallic scraps, paper wastes and some plastics can be recycled.
b) Incineration.
It involves burning the wastes. The heat can be used for heating. It kills germ. However, the
fumes and other products of incomplete combustion can pollute the environment.
c) Land-fill:
Wastes are covered with a thin layer of soil. The wastes get compacted together. The layer
has to be of reasonable size otherwise it may get corroded during the rains and expose the
wastes to the environment.
d) Composting:
Turning wastes into organic manure.
e) Production of bio-degradable materials:
This involves manufacturing and using material that can easily be broken by the action of
micro-organisms.
HOW THE TOPICS (ORGANIC CHEMISRTY 1 & 2) HAVE BEEN FEATURED IN THE LAST 4 YEARS
OF MANEB EXAMINATIONS
(Find some suggested answers in italics).
110
2015 3a
i. Name the products formed during the fermentation of sugar by yeast
Answer: Ethanol and carbon dioxide
ii. Describe how fermentation of sugar by yeast occurs
Answer: the yeast contains natural enzymes which catalyze the breakdown
of glucose to ethanol and carbon dioxide. Heat increases the rate at which
fermentation is done.
b. Explain why thermosetting plastics do not melt when heated
Answer: their polymer chains are joined together by cross-links, so they cannot slide or glide
past each other when heated
c. The following are isomers of organic compounds A and B
(i) Name the isomers A and B
A: is 3-methybut-1-ene B: cyclopentane
(ii) To which homologous series do the isomers belong?
A: belongs to alkenes B: alkanes
d. Polymerization of ethane takes place according to the following equations
(i) Name the polymer Answer: Polythene
(ii) Give any two advantages of the polymer
It cannot be attacked by microorganism
Can be recycled
It is light
(iii) Explain how the polymerization takes place
The monomers are exposed to conditions that allow the carbon-carbon double bond to break
giving room for formation of a long chain
111
2014 No 5
a. What are hydrocarbons?
These are organic compounds with carbon and hydrogen atoms only
b. Figure below shows formula of some organic compounds A, B, C and D
A
OHHC
73
B
HC
105
C
COO HHC
52
D
HC
209
(i) Which compound are hydrocarbons?
Answer: B and D
(ii) Name compound A
Answer: Propanol
(iii) Draw molecular structure of compound C
Answer
(iv) Complete the flow diagram for differentiating compounds B and D by filling in the
missing information in the brackets
Answer
c. State any three properties of synthetic polymers
Are very strong
Do not conduct electricity
Cannot be decomposed by micro organism
112
Are versatile (have many uses)
d. Draw the structures of the two isomers of butane(
104
HC
)
2013 N0 5
a. Figure below shows structures of some organic compounds W, X, Y and Z
Note: Compound X was not correctly drawn in Maneb exam. The double bond was not
supposed to be there.
(i) To which homologous series do compounds X and Z belong?
X: alkanols
Z: alkenes
(ii) Mention any three chemical properties of compound X
React with carboxylic acids to form esters
Undergo combustion to produce energy
React with sodium to form sodium alkoxide
Gets oxidized to carboxylic acid
(iii) Write the general formula for compound Y
COOHHC
nn 12
(iv) Name two products that are formed when compounds X and Y react
Esters and Water
(v) Which compounds belong to the same homologous series?
W and Z
(vi) Give a reason for your answer above
They have the same functional part; the carbon carbon double bond for
alkenes
b. Figure below shows structures of plastics A and B
113
(i) Which structure represents thermosetting plastics?
Answe:r B
(ii) Give a reason for your answer above
Answer: It has cross-links which make it rigid, among other properties
c. Explain how ethanol (
OHHC
52
)could be distinguished from hexane (
)
146
HC
By solubility test. If a few samples of each are mixed with water, ethanol is soluble;
forms one layer while hexane is insoluble; it forms two layer
2012
3 .The following are general formulae of organic compounds A and B
A = C
n
H
2n+1
COOH B = C
n
H
2n+1
OH
a. To which family does compound B belong? Alkanols
b. Mention any three properties of compound A
I. Long chain acids are liquids at room temperature.
II. Melting and boiling points increase with an increase in molecular mass.
III. Are soluble in water. However, solubility decreases as number of carbon atoms in the
molecule increases.
IV. They react with carbonates and hydrogen carbonates to form salts and rebate carbon
dioxide.
OHCOCOONaCHCONaCOOHCH
223323
2
V. They neutralize alkali to form corresponding salts and water.
OHCOOKCHKOHCOOHCH
233
VI. Their solutions affect color of indicators.
VII. They react with alkanols to form esters
c. State any three uses of compound B
i. Used as a source of fuel.
ii. Used as essential ingredient of alcoholic beverages.
iii. Used as a raw material in pharmaceuticals, perfumes and flavorings.
iv. Used as a solvent for many organic substances.
v. It is an intermediate in the manufacture of other chemicals such as acetaldehydes.
vi. Used in ester production
d. Mention the products formed when compound A and B react
Ester and water
e. Work out the molecular formula of compound A if n is 5
114
C
5
H
11
COOH
f. Describe how compound A could be distinguished from compound B
By carrying out an acid test. Add a few drops of dilute sodium hydroxide, NaOH
in two separate test tubes and add 1 or 2 drops of phenolphthalein indicator. It
turns pink.
Add chemical A into one of the test tube. A color-less mixture will indicate it’s
an acid. Pink will signal presence of B
8. ELECTRICITY AND MAGNETISM 1
OBJECTIVES:
By the end of the chapter learners should be able to:
Explain the different types of electricity.
Analyze types of charges
Describe the uses of electrostatics in our everyday life.
Discuss current electricity.
Calculate total resistance in series and parallel circuit.
Deduce resistance of resistors from color codes and standard notation.
Explain the meaning of electrical power.
Calculate the electrical power using different equations.
Interpret power rating of electrical appliances
Calculate total electrical energy used.
Determine the cost for electrical energy.
Describe method of magnetization and de-magnetization of steel or iron.
Discuss the phenomenon of electro-magnetism.
Describe the interaction of current and magnetic field
Determine the field pattern for a conductor currying current.
Investigate the induction of emf.
Explain how a transformer works.
Use transformer equations in problem solving.
Describe power loss and control in transformers
STATIC ELECTRICITY:
There are two types of electricity namely static electricity and current electricity.
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Static electricity is the type of electricity which does not involve charge movement.
The term charge refers to electrons and ions.
There are two types of charges and these are positive charges and negative charges.
Electrostatics is usually experienced in insulators in which the outer most electrons are
strongly held by their nuclei.
Between or among charges there exist forces of attraction and repulsion.
The law of charges state that like charges repel and unlike charges attract.
A material with the same number of charges is said to be neutral.
When a cellulose acetate strip is rubbed with a piece of white cloth or against hair, it
becomes positively charged. The polythene strip becomes negatively charged when
rubbed by the same.
The charging result into in-balance of the number of electrons. There are more positive
charges in a rubbed acetate strip than the negative charges while polythene strip has
many electrons than protons.
The number of electrons exchanged determines the degree of strength of the charges.
The electrical charge is measured in Coulombs.
METHODS OF CHARGING:
An object can be charged by the following methods:
Friction (Rubbing).
By contact.
Induction.
Charging by rubbing causes electrons to jump from the material into cloth/hair (in case
of cellulose acetate strip) leaving the material positively charged. Electrons, sometimes,
move into the material being rubbed such that it is left with excess of electron leaving it
negatively charged like polythene strip.
Charging by contact: when a charged material is brought closer to a neutral one,
separation of charges takes place. The objects are then brought in contact, the unlike
charges cancel each other out (discharge). Refer to the steps below
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Charging by induction does not involve touching of objects. There is separation of
charges. An earth wire provides path for electrons into the ground. A person may just
touch the object and the electrons would move through the person into the ground.
Follow the figure below
Materials which normally get positively charged include cellulose acetate strip, glass,
and Perspex.
The list of materials that become negatively charged when rubbed include polythene
strip, balloon rubber, Bakelite and silk.
Note that metals cannot be charged by rubbing because the charges are conducted
away as soon as they are gained so that the balance of negative and positive charge is
restored.
CHARGE DETECTION:
The presence, type and strength of an electric charge can be determined by
electroscope. This is a device which detects charge.
Most commonly used electroscopes are Pith ball electroscope and Gold leaf
electroscope.
a. Pith ball electroscope:
o This is a very light ball that has a conducting surface of aluminum and is suspended
freely by a thin insulating thread of nylon.
o A rod of known charge is brought in contact with the pith ball and repulsion reveals that
the pith-ball and the rod are of the same charge.
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o Alternatively, a rod of unknown charge is brought near to the pith-ball electroscope of
known charge. Attraction suggests the two possess opposite charges.
b. Gold-leaf electroscope:
o Suppose the cap and the leaf are negatively charged and a negatively charged body is
brought near the cap, then the electrons will be repelled down the leaf and the leaf will
diverge further.
o In general if the leaf of a charged electroscope diverges further when an object is placed
near the cap, the object is charged and the charge has the same sign as that on the
electroscope. The underlying rule is that like charges repel while unlike charges attract.
LIGHTNING:
This is a giant spark of electricity.
IT happens due to friction between raindrops and ice crystals caused by wind. As they
rub together violently, friction causes the transfer of electric charges (electrons) from
one drop to another. As the amount of opposite charges build up the force between
them and the bottom clouds becomes greater and greater. This results into occurrence
of the giant spark of electricity (lightning) coupled by a heavy sound (thunder) due to a
sudden expansion of air. This takes place when the electricity passes through air.
When a conductor is charged up, the charges repel each other so they collect on the
outside; the charges are concentrated near the sloppiest points. This is where the
electric filed is strongest and the field lines closest together.
PREVENTION OF EFFECTS OF LIGHTENING BY A CONDUCTOR:
A lightning conductor which is usually made from a strip of copper has a sharp point.
The conductor does two functions:
a. Providing path for the lightning to the ground.
b. Discharging the cloud there by preventing lightening from striking.
Normally, if a positively charged cloud passes over a building conductor a negative
charge is induced on the point at the top of the cloud. Because of high density charge at
the point, electrons are sprayed into the air and attracted by positive charge on the
cloud and make a lightening spark less likely to occur. If a discharge does take place, the
charges flow harmlessly down the lightening conductor.
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APPLICATIONS OF ELECTROSTATICS:
There are several applications of the knowledge and concepts of electrostatics some of
which will be discussed.
a. Designing of Capacitors:
o A capacitor is a device which is used to store electric charges.
o A simple capacitor consists of two flat pieces of metal separated by an insulator
of very high resistance called dielectric.
o The ability of a capacitor to store charge is measured by its capacitance.
o Thus, capacitance is the amount of charge stored on each plate of capacitors per
unit voltage.
o Capacitance are used for:
Smoothening out current changes.
Passing on signals from one circuit to another.
Tuning circuits so that they respond to signals of one frequency e.g.:
radio-tuning.
Used in flash cameras to store charge.
b. Functioning of Photocopiers:
o Photocopiers contain a negatively charged drum and when the paper to be
copied is laid on the glass plate, the light reflected from white parts of paper
causes the charge to disappear from the corresponding parts of the drum. The
charge pattern remaining on the drum corresponds to the dark colored printing
on the original. Powdered ink (toner) is then dusted over the drum, and sticks to
the parts which are still charged. When a sheet of paper passes over the drum,
the particles of toner are attracted to it and fused into place by a short burst of
heat. The heat melts the powdered ink to enable it stick to the paper.
c. The working of electro-statics Precipitators:
o This is a way of cleaning up pollution of air in the form of smoke (dust particles)
from the atmosphere.
o The smoke is passed through electric field which produces ions which are
adhered to the dust particles giving the charge.
o The charged particles are attracted towards an earthed plate and collect on it.
o They are periodically removed by striking the plate with a mechanical hammer.
The dust particles fall into collector for disposal.
d. Paint Spraying:
o In paint spray guns, the static charge ensures an even distribution of the drop-
lets of paint. The same happens in crop sprayers.
CURRENT ELECTRICITY:
Current electricity is the uniform flow of electric charge. It is usually the free uniform
movement of electrons.
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A circuit is the path traced by the electrons. Circuit symbols are used when drawing a
circuit.
There are two types of circuits namely: series circuit and parallel circuit.
In series circuit, components are joined in a single conducting loop (path).
In parallel circuits, components are joined in several conducting paths.
The following are some of the circuit symbols:
Conventional current flows from positive to negative terminals.
A terminal current is considered to flow from negative to positive.
Movement of charges requires energy. This is the Voltage or the potential difference.
Thus, Voltage is the amount of energy or work required to move a unit charge between
two or more points.
Electro-motive force, EMF is the potential difference across the terminals of a cell or
battery when not connected to a circuit. It is usually labeled on the cell, say 1.5V.
In any circuit, there are three things which have to be determined. These are
summarized in the table below:
QUANTITY
ITS SYMBOL
SI UNIT
MEASURED BY
CURRENT
I
AMPS, A
AMMETRE
VOLTAGE
V
VOLTS, V
VOLTMITRE
RESISTANCE
R
OHMS,Ω
OHMMETRE
ELECTRICAL RESISTANCE:
It is the opposition to the flow of electric current.
It is due to the ability of the substance to slow down the uniform speed of charges in an
electric circuit there by reducing current size.
When electrons are flowing they collide with stationary ions causing the ions to vibrate.
As the electrons are colliding with the ions, they get slowed down.
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The degree of resistance varies from material to another such that there are some with
low resistance and others with high resistance.
Good conductors of electricity have low resistance because they have free electrons less
strongly held by the ions. Examples of good conductors of electricity are copper, silver,
gold etc. These are used as connecting wires for electricity.
Materials like nichrome (nickel + chromium), tungsten and constantan (60%
copper&40% nickel) have high resistance. The higher the resistance the higher the value
of heat energy produced.
Nichrome is used as heating element of electric kettles, ovens and geezers.
Tungsten is used in bulb filament and constantan is used in designing of resistors.
RESISTORS:
These are conductors which provide the opposition to flow of current.
They are used to:
o Limit flow of current there by protecting appliances from excessive
currents.
o Control volume in radios (in case of variable resistors).
o Produce heat.
Ammeters do have a low value of resistance so that they measure the exact current
value passing through an appliance. ON the hand, voltmeters do have high resistance
hence are connected in parallel with the component so as not to reduce current value.
The voltage, V, current, I and resistance, R are related by the equation
IRV
. This
equation is derived from the Ohm’s Law which states that “For a metal conductor at
constant temperature, the potential difference (P.d) applied across the ends of the
conductor is directly proportional to the current I in the conductor.
Thus,
IV
Graphically,
Voltage, V
I
V
R
Current, I
The slope of the above graph gives the resistance as show.
The data for the graph can be verified by setting up an apparatus as illustrated in the
figure below
121
.
Coil of resistance wire. (Can be a variable resistor, which is also called Rheostat).
Voltage is varied by adding cells while resistance is varied by using different lengths of
the same wire of uniform cross section.
FINDING TOTAL (EFFECTIVE) RESISTANCE:
RESISTORS IN SERIES:
The total resistance for resistors in series is obtained by finding the sum of the total
resistances.
R
1
R
2
R
3
Total Resistance= Sum of the resistances.
I.e.
321
RRRR
T
Example: Calculate the total resistance in the following circuit:
R
T
= 5Ω + 2Ω + 7Ω = 14Ω
RESISTORS IN PARALLEL:
The resistance of the resistors in parallel is found by the following equation:
nT
RRRRR
1
..........
1111
421
If only two resistors are used in parallel then the total would be given by the following:
21
111
RRR
T
sum
oduct
RR
xRR
R
T
Pr
21
21
Example: Work out the total resistance in the following:
5Ω
7Ω
2
Ω
122
67.2
12
32
48
48
2
21
21
x
RR
xRR
R
T
If two resistors of equal resistance are joined in series then the total resistance would
just be twice the value of resistance of one of the resistors.
E.g. Given that an 8Ω resistor is joined in series with another 8Ω resistor, find the total
resistance. R
T
= 2 R
1
= 2 R
2
= 2 x 8Ω = 16 Ω
On the other hand, if two resistors of equal resistance are joined in parallel then the
total resistance is just half the resistance of one of the resistors.
Given that the same resistors above are joined in parallel to each other what would be
the total resistance? R
T
= (8 x 8)/ (8 +8) = (64 /16) = 4Ω which is just half of 8Ω.
INTERNAL RESISTANCE OF A CELL:
Practically, it is found out that chemicals inside a cell have a small resistance of their
own. This is called internal resistance of a cell. This implies that some voltage (potential
difference) gets used up inside the cell.
Denoting the internal resistance by r one would have the following equation:
VvIRIrE
I
VE
r
.
E is electro-motive force (emf).
Refer to the figure below
The cell, in the figure above has an emf of 1.5V.The voltmeter reading is 1.35V.
Ammeter records the value of the current as 0.3A. Calculate (a) the internal resistance
of the cell and (b) the value of the resistance R.
a)
5.0
3.0
35.15.1
A
VV
I
VE
r
123
b) Since E = Ir + IR then
5.4
3.0
5.03.05.1
A
AxV
I
IrE
R
.
FAST FACTS:
1) Current in series circuit is the same
2) Voltage in series circuit is shared across components
3) Current in parallel circuit is shared at junctions
4) Voltage across components in parallel circuit is the same
ADVANTAGES OF PARALLEL CIRCUIT OVER SERIES CIRCUIT
a) Components work independently
b) If a component is faulty, the other components still work
c) In case of bulbs, addition of more bulbs has no effect on brightness of the others
CIRCUIT PROBLEMS
1. The fig below shows a circuit
Work out the following
a) Ammeter reading
b) Voltage across each of the resistors
a) R
T
= 1Ω+2Ω=3Ω so
A
V
R
V
I 5.0
3
5.1
b) Across 1Ω resistor V = IR= 0.5A x 1Ω =0.5V and Voltage across 2Ω resistor =IR =
0.5A x 2Ω= 1V
2. Use the circuit below
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Work out the ammeter reading when
a) S1 is open and S2 is closed
b) S1 closed and S2 open
c) Both S1 and S2 are closed
a)
A
V
R
V
I 3
)32(
15
b)
A
V
R
V
I 5.1
10
15
c)
15
50
510
510
)32(10
)32(10 xx
R
A
V
R
V
I 5.4
15
50
15
3. If the total resistance is 4Ω, find the value of resistance x in the set up below
Ans: 5Ω
4. Find the total resistance in the following circuits
e.
Answer : 2.1Ω
f.
Answer: 12Ω
g.
Answer: 4Ω
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FACTORS THAT AFFECT THE RESISTANCE OF A CONDUCTOR (WIRE):
There are four factors that affect the magnitude of resistance in a conductor and these
are: length, temperature, thickness of the wire and type of material from the
conductor is made.
a) Length:
The longer the wire the higher the resistance and vice-versa. IF the wire is longer, the
electrons bump or collide into each other for a long distance hence having a huge effect
on the resistance. This consequently, reduces current flow. Thus, the resistance and the
length of wire are directly proportional to each other but both inversely proportional
to the current.
Resistance
(Ohms)
Length (cm)
b) Temperature:
The higher the temperature, the higher the value of the resistance. This is attributed to
the fact that increased temperatures cause rapid vibration of ions within the wire. This
results into more collisions with the electrons. However, temperature increases
decreases the resistance of semi-conductors.
c) Thickness (Cross-sectional area):
The thicker the wire the lower the resistance and the thinner the wire the higher the
value of resistance. Thus, the thickness of a conductor is inversely proportional to the
resistance. This is so because in a thicker wire there is more space for electrons to pass
through hence electrons experience minimal collisions.
d) Type of material from which the wire is made:
Materials do vary in how they oppose flow of electrons. For instance, copper has low
resistance while nichrome has high resistance. This property determines how best the
materials are used.
COLOR CODES FOR RESISTANCE:
This involves use of colored bands to indicate value of resistance.
The first color gives the first digit, the second color second digit, third color is a
multiplier (it gives number of zeros) and the forth digit if any, gives tolerance.
THE tolerance refers to how the actual value differs from the given value of resistance.
The following are some figures of tolerance:
None =
20% Silver =
10%Gold =
5%Red =
2%Brown =
1%.
The colors black, brown, red, orange, yellow, green, blue, violet, grey and white
represent digits from 0 to 9 respectively.
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The colors are encompassed in the following statement: “Black Boys Raped Our Young
Girls But Violet Gave willingly.”
COLOR
Black
Brown
Red
Orange
Yellow
Green
Blue
Violet
Grey
White
DIGIT
0
1
2
3
4
5
6
7
8
9
For example:
a. Give the resistance represented by the color code on the following resistor
Red, Yellow, Red, Silver
Answer: 2400Ω
10%
b. Use color codes to represent the resistance of a resistor below:
460000Ω
5%
Answer: Yellow, Blue, Yellow, Gold.
PRINTED CODES (STANDARD NOTATION):
Letters R, K and M are used. R is for unit, K for kilo or thousand and M for mega or
million.
The position of the letter signals presence of a decimal point.
Tolerance values are: F =
1%, G =
2%, J =
5%, K =
10% and M =
20%.
Examples:
PRINTED CODES
RESISTANCE VALUE
7R6
7.6Ω
6K2J
6.2KΩ
5%= 6200Ω
5%
R8G
0.8Ω
2%
4M8K
4.8MΩ
10%= 4.8 X 10
6
Ω
10%
M7
0.7MΩ = 7 X 10
5
Ω
3RF
3Ω
1%
ELECTRICAL POWER:
This is the rate at which electrical energy is transferred to other forms of energy.
It can also be defined as the rate of using up electrical energy.
Thus, power = (Work done)/(Time) = (Energy transferred)/ (Time(taken).
The SI unit of power is Watt. 1 Watt = 1J/s. 1KW = 1000W.
Power = Voltage x current = VI
If a bulb has a power rating of 75W, it means the bulb as a transducer, converts 75
Joules of electrical energy into heat energy and light energy.
Since P = VI and from Ohm’s Law V = IR then P = (IR) I = I
2
R and P = V (V/R) = V
2
/R
Given Voltage and Current
VIP
Given Current and Resistance
RIP
2
Given Voltage and Resistance
R
V
P
2
Examples.
127
a. Calculate the maximum power of an electrical appliance that can be connected
safely to 13 A 240V mains socket.
Given I = 13A and V = 240V P = VI = 13A x 240V = 3120W = 3.12 KW.
b. A 3KW electric fire is designed to be run on a 250V supply. Assuming it is operating
at a correct voltage calculate the current it will draw from the mains and the
resistance of the element of the fire.
o Given P = 3KW = 3000W and V =250V.
P =VI then I = P/V = 3000W/250V = 12A.
o P = V
2
/R then R =V
2
/P = 250
2
/3000 = 20.8Ω or use V = IR then R = V/I =
250/13 = 20.8Ω
COST OF EECTRICAL ENERGY:
The electricity going into a consumer destination passes through a kilowatt- hour meter.
This meter measures and records the electrical energy a customer uses in Kilowatt-
hours.
A Kilowatt-Hour is the total electrical energy that is spent for one hour at a constant rate
of 1KW or 1000W of power.
For example a 1000W heater turned on for one hour would consume 1KWh of electrical
energy. A 100W lamp left for 10 hours would also consume 1KWh of the electrical
energy.
The 1KWh of electrical energy is called a “unit”
Example: A fridge with a power rating of 150W operates for 12 hours. Work out the cost
of using the electricity if 1 unit of electrical energy costs K10.
Electrical energy = power (KW) x time (hrs) = (150/1000) KW x 12 hrs = 1.8KWh.
Therefore, there are 1.8 units of electrical energy. Since 1 unit cost K10 then 1.8 units
will cost more. The Cost = 1.8 units x K10/unit = K18.00.
SAFETY OF USING ELECTRICITY:
It is recommendable to use three pin plugs when using power.
A three pin pug has a live pinconnected to a live wire, earth pin connected to an earth
wire and a neutral pin which is connected to a neutral wire.
A live wire has highest potential difference, carries current and is connected to a fuse. It
is brown in color (old color red)
A neutral wire is at 0V and it just completes circuits. It is blue in color (old color black)
An earth-wire is for safety purpose. It is green yellow (old color green). It is connected to
a metal chassis of an appliance and the other end is buried in earth. It provides a
passage for electrons into the earth in case of short circuits. The earth is a bank of
electrons.
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A fuse is a small metal alloy which has a low melting point. It breaks circuit when too
much current flows through it hence protecting appliances from damage.
Fuse rating is the description of the highest current value a fuse can allow before
melting. Many available fuses have a fuse rating of 13A, 3A, 1A, 10A etc.
MAGNETISM:
Magnets are objects which have the ability to attract other objects. Usually these
objects contain iron or steel.
Materials that get attracted by magnets are Ferro-magnets. Examples of Ferro-magnets
are cobalt, nickel, gadolinium, dysprosium etc.
Magnetic poles:
These are places in the magnets to which magnetic materials are attracted. They are
found near the ends and they occur in pairs of equal strength.
Magnetic attractions are stronger in the poles.
North Seeking pole (N-pole): This always points roughly towards the Earth’s North Pole.
South Seeking pole (S-pole): This always points roughly towards the Earth’s South Pole.
Like poles repel and un-like poles attract each other.
Types of Magnetic materials:
a. Soft materials:
o These are materials that are magnetized easily but they also lose their magnetism so
easily e.g. soft iron.
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o Magnets produced by soft materials are temporary.
b. Hard materials:
o These are materials that get magnetized slowly and slowly they lose their magnetism
e.g. hard steel.
o Magnets produced by hard materials are permanent.
o Magnetism that get induced in iron is temporary where as that which gets induced in
steel is permanent.
o Magnets can be of different shapes and strength depending on the use. There are bar
magnets, horse shoe magnets, loud speaker magnets etc.
Magnetic field:
This is a region around a magnet where a magnetic force is experienced by an object
placed in the region.
The magnetic field is represented on paper by a set of lines which are referred to as field
lines.
The density of the field lines indicates the intensity of the magnetic field.
Magnetic field lines can be mapped by using iron filings or using a small plotting
compass
Note the following facts:
a. The pattern of each side is symmetrical.
b. The magnetic field lines seem to originate from the North Pole to the South Pole.
c. The field lines are more concentrated at the poles
d. The magnetic field lines seem to come together at the poles (but not right at the
ends).
Examples of magnetic field diagrams:
MAGNETIZATION (MAKING OF MAGNETS):
a. By Electrical Method:
This involves placing a magnetic material in a solenoid. A solenoid is a cylindrical
coil of wire carrying current.
Magnets which can be controlled by the flow of electric current are called
electromagnets. Their magnetism is temporary and is used in places where
strong magnets are needed and permanent magnets would not be suitable.
Materials which are frequently used in solenoids of electromagnets are iron and
nickel alloy which lose their magnetism once current is switched off.
The Polarity (which side is north or south) can be determined by looking at how
the coil has been hooked.
130
The right hand grip rule predicts which end will be the North Pole.
Reversing the current direction reverse the polarity of the magnet.
The Strength of the electromagnets can be affected by the following:
i. number of wire coils:
The higher the number of wire coils the stronger the magnet.
ii. Size of the current:
The larger the current size the stronger the magnet.
iii. Distance between the poles:
The closer the poles are, the stronger the electromagnet.
b. By Stroking Method:
There are two stroking methods namely;
i. single stroking
ii. Double (divided) stroking.
I. Single stroking:
Stroking means making gentle caressing movements.
This method involves moving one bar magnet repeatedly, in the same direction
above an object to be magnetized.
The polarity of the yet to be produced magnet would be in a way that the end
will have a pole opposite to the pole near it.
If a magnet is snapped into half, both halves will have north and north seeking
poles and would each be a complete magnet.
131
II. Double stroking:
This is done by stroking two magnets from the Centre outwards with unlike poles
of the magnets at the same time. The magnet s must be lifted high above the
steel at the end of each stroke in both methods.
The pole produced at the end of the steel where the stroke ends is opposite to
that of the stroking pole.
Note:
The above methods technically, cause alignment of small magnets within a magnetic object.
These are called domains. In non-magnetic materials these domain are not well arranged. The
magnetism effect is canceled out.
DEMAGNETIZATION OF MAGNETS:
This entails that magnets can lose their magnetism through different ways. Some of them are
discussed.
a) Heating:
This results into the destruction of the alignment of the tiny magnets (domains).
b) Hammering (hitting):
This also causes de-arrangement of the domains inside the magnets.
c) Use of alternating current:
This involves placing the material inside a solenoid through which an alternating current is
flowing. When the current is slowly reduced to zero, the magnet becomes de-magnetized.
d) Stopping current flow:
132
One, needs electricity for electro-magnets to be produced hence stoppage of current flow
leads to loss of magnetism.
The knowledge on how magnets lose their magnetism is vital because one is able to figure out
ways of taking care of the magnets. Normally, the magnets have to be kept in a wooden box,
and avoid dropping the magnets down.
USES OF ELECTROMAGNETS:
a) Electric bell:
o The soft iron cores are wound so that they have opposite polarity. The soft iron
armature is attached to the steel of the spring.
o When a switch is on, the charge flows; the soft iron cores become magnets and attract
the armature, ringing the bell in the process.
o As the contact is broken the armature moves down the magnet at C and the current is
automatically switched off. The soft iron bars lose their magnetism and the armature is
no longer attracted. The steel spring causes the armature to spring back and remake the
contact at C. This switches on the current and the cycle is repeated.
b) Magnetic relay switches
This device uses one circuit to switch on another. They may also be used to control
circuits carrying the currents. This operation hugely depends on the magnetism.
c) telephone ear piece
d) loud speakers
e) Lifting of scrap metals.
FORCE ON A CONDUCTOR IN A MAGNETIC FIELD:
When an electric current flows through a wire or a cable a magnetic field forms around
the wire.
The field or flux pattern around a current carrying wire is a series of concentric circles.
The direction of the magnetic field can be found by using the Right Hand Grip rule. I.e. if
the fingers of the right hand grip the wire with the thumb pointing in the direction of
current, the curl of the finger show the direction of the magnetic field.
133
The direction of the field can also be predicted by the Maxwell’s right handed screw
rule. “If a right handed screw moves forward in the direction of current
(conventional), the direction of rotating of the screw gives the direction of the filed.
The field can be made stronger if the wire is formed into a coil ( solenoid)
When a current carrying wire is placed in a magnetic filed it experiences a force.
The magnitude of the force depends on two factors:
I. Field strength:
As the magnetic field strength increases, the force also increases.
II. Current size:
As the current increases so too does the force.
The direction of the force (which is also the direction in which the wire moves) depends
on the direction of the current and direction of the magnetic field.
The direction of the force can be predicted by FLEMING’S LEFT-HAND RULE (MOTOR
RULE).
The above Left hand motor rule states that “ Hold the first finger, second finger and
the thumb of the left-hand mutually at right angles so that the First finger points in
the direction of the Field, the Thumb in the direction of motion (Thrust) then the
second finger in the direction of current.”
Right hand grip rule (dynamo rule)
134
The rule above gives the direction of induced current.
ELECTRIC MOTOR:
In a motor current passes through the coil and the resulting force on the coil causes
it to rotate. The direction of motion is determined by Fleming’s Left hand Rule.
A split ring (commutator) reverses the direction of current every half the revolution,
thus ensuring a continuous rotation.
The size of the turning couple on the armature of a motor may be increased in the
following ways.
I. Increasing the magnetic field strength by using stronger magnets.
II. Increasing the number of turns on the coil
III. Increasing the current size.
IV. Increasing the area of the coil.
The largest couple act when the coil is moving at right angles to the magnetic field. The
greater the couple, the faster is the turning of the motor.
In electric motor soft materials such as carbon or phosphor bronze “brushes” are used
to make contact with the commutator segments as they rotate.
The brushes tend to wear out with time much more quickly than the harder copper of
commutators and as such they must be replaced.
ELECTRO MAGNETIC INDUCTION:
When loop of wire is moved at right angles through a magnetic field a potential
difference is created across the ends of the wire and small current flows.
The potential difference produced in this way is called induced emf and the current is
called induced current.
Electromagnetic induction can be defined as production of electric current using
magnetism.
Factors that affect the size of induced emf and current:
i. Speed of the wire coil:
The faster the coil moves the greater the emf and current induced.
ii. Strength of the magnetic field:
135
Stronger fields produce large emf and currents.
iii. Number of turns in the coil:
Increasing the number of wire passing or cutting through the magnetic field increases
the emf and the induced current.
iv. Area of the coil:
Increased area of the coil produces high emf and current.
USES OF ELECTROMAGNETIC INDUCTION:
The ability to produce an emf by changing the magnetic field can be used in different
ways like production of electricity using generators and in transformers.
TRANSFORMER:
It is an electrical device by which alternating current of one voltage is changed to
another voltage.
The circuit symbol of transformer is
HOW A TRANSFORMER WORKS:
An alternating current is fed into the primary coil and it creates a magnetic field in it.
The alternating magnetic field cuts through the secondary coil generating an induced
alternating voltage across the secondary coil. The current from the secondary coil is
therefore an alternating current.
As the secondary voltage is made bigger the secondary current gets smaller because the
transformer can increase voltage but not power.
Alternating current (a.c) means current that changes direction of flow. (It goes negative
then positive then negative and so on.) The opposite, is direct current, dc which is a one
way current.
Soft Iron Core
ac
Primary coil Secondary coil
TRANSFORMER EQUATIONS:
1. This relates the ratio of voltages and number of turns:
136
p
s
p
s
N
N
V
V
Where
V
s
Voltage in the secondary coil
V
p
Voltage in the primary coil
N
s
Number of turns in the secondary coil
N
p
Number of turns in the primary coil
2. It gives relationship among currents and voltages in the secondary coil and primary coil
of the transformer.
s
p
p
s
V
V
I
I
Where I
s
is the current in the secondary coil and I
p
is the current in the
primary coil.
3. Assuming that the transformer is ideal ( there is no any power loss) then
Power in = Power out or Power in the primary = Power in the secondary coil.
Since P =VI then
sspp
xIVxIV
Efficiency of a transformer refers to how well a transformer works. If loss of power is
zero the transformer is ideal or perfect otherwise power loses are inevitable. Usually the
power in the primary coil is greater than the power in the secondary coil.
Efficiency = (Power output)/ (Power input) = (V
s
I
s
)/ (V
p
I
p
).
%efficiency =
%100x
IV
IV
pp
ss
TURNS RATIO: This is the ratio of number of turns in the primary coil to the number of turns
in the secondary coil
Thus Turns ratio = N
p
/N
s
.
TYPES OF TRANSFORMERS:
There are two types of transformers which are step up transformer and step down
transformer.
The Step-up transformer has more turns in the secondary coil than in the primary coil. It
also has higher secondary voltage than the primary voltage.
The Step-down transformer has more turns in the primary coil than in the secondary
coil. In addition, there is a larger primary voltage than secondary voltage.
Examples:
1. A transformer with 2000 turns in its primary coil has a primary voltage of 120V. If there
are 6000 turns in the secondary coil, work out the secondary voltage, assuming that the
transformer is 100% efficient. What type of transformer is this?
137
Solution:
N
p
= 2000 V
p
= 120V N
s
= 6000.
p
s
p
s
N
N
V
V
VVxxV
N
N
V
p
p
s
s
360120
2000
6000
Type of transformer: Step-up.
2. a. Calculate the number of turns of the secondary of the step down transformer, which
would enable a 12V bulb to be used with a 240V a.c mains power, if there are 480 turns
on the primary.
b. What current will flow in the secondary when the primary current is 0.5A? Assume
there are no energy loses. (Duncan T, 1977 page 211; 2
nd
Ed.)
Solution:
a. N
p
= 480 V
s
= 12V, V
p
= 240V
p
s
p
s
N
N
V
V
24480
240
12
xxN
V
V
N
p
p
s
s
Thus, there are 24 turns in the secondary coil.
b. I
p
= 0.5A
sspp
xIVxIV
A
x
V
IV
I
s
pp
s
10
12
5.0240
3. A 240V mains transformer has an efficiency of 90% and is used to light normally a 12V-
36W lamp. Work out
a. The power in the primary coil.
b. The current in the primary coil.
Solutions:
a. V
p
= 240V, V
s
= 12V P
s
= 36W %E= 90%.
%90%100
36
%100% x
P
W
x
P
P
E
pp
s
W
Wx
P
p
40
%90
%10036
.
b. P = VI
P
p
= V
p
I
p
therefore I
p
= P
p
/ V
p
= 40W/240V = 1/6 =0.17A.
ENERGY LOSSES IN TRANSFORMER:
a. Copper windings:
o The windings of copper wires have some resistance and as such some heat is
produced by the currents in the wires.
o Use of properly designed transformers may help in minimizing the energy
loss in this form.
138
b. Eddy current loses:
o Eddy currents are currents due to fluctuation of magnetic field. The increase
and decrease of the magnetic field results in the production of this eddy
current since the core is also a conductor. These currents heat the core and
energy is lost in the process.
o To reduce the effect of eddy currents the core is made from thin layers of
soft iron glued together with an insulating material between the strips.
o So, eddy currents occur whenever pieces of metal are in a changing magnetic
field e.g. transformer.
c. Flux leakage:
o All the field lines produced by the primary coil may not ‘cut’ the secondary
coil especially when the core has air gaps or has been poorly designed. This
loss of field lines can be minimized by inserting the gap with iron core and
ensuring that the transformer is properly designed.
d. Magnetization and demagnetization of the core:
o Using soft iron for the design of the core reduces the losses to minimum.
TRANSMISION OF ELECTRICAL POWER:
Electricity is transmitted at high voltage (high turns) and at low current so as to reduce
the heating of cables through resistance.
This result form the fact that power is given by the equation P = I
2
R.
A transformer is used to step up the power station voltage to over a quarter of a million
volts. This makes the current smaller. The voltage is stepped down again when it
reaches consumer points in towns and villages.
DIFFERENCES BETWEEN DIRECT AND ALTERNATING CURRENT:
1. In direct current (d.c), electricity always flows in one direction (along the same path)
while in alternating current (a.c) the current direction changes regularly.
2. Cells and batteries provide d.c while the mains supply is a.c, alternating at frequency of
50 cycles per second (50Hz).
ADVANTAGES OF AC OVER DC:
1. Voltages can be stepped up or down using transformers.
2. AC is easier and cheaper to generate than DC.
3. On a large scale it can be generated more efficiently than the one way DC.
139
9. ELECTRICTY 2 (ELECTRONICS).
OBJECTIVES:
By the end of this chapter learners should be able to:
Explain the meaning of band theory
Explain the meaning of semiconductors
Discuss uses of semiconductors (diodes, transistors etc).
BAND THEORY:
When large numbers of atoms are together as they are in a crystal, the energy levels
spread into bands.
Each band contains a large number of levels and these are so close together that in
effect, there is a continuous range of energies available to the electrons.
The energy bands are separated by gaps in which there are no available energy levels.
If an electron is to take part in the conduction of an electric current it must be capable
of being accelerated by an applied voltage and so must be capable of being raised to a
slightly higher energy level.
A material can therefore conduct electricity only if some of its electrons are in a band
which does not contain its full quota of electrons.
SEMICONDUCTORS:
These are materials which conducts electricity partially.
Their conductivity is higher than that of insulators and less than those of conductors.
Examples of semiconductors include Silicon (Si), Germanium (Ge), Antimony (Sb),
Indium (In), Arsenic (As) etc.
Electrical resistivity of semiconductors decrease with an increase in temperature.
Insulators and semiconductors differ only in that all insulators are semiconductors at
high temperatures and all semiconductors are insulators at low temperature.
The conductivity of semiconductors is affected by the presence of impurities.
TYPES OF SEMICONDUCTORS:
1. Intrinsic semiconductors:
These are semiconductors whose conductivities arise due to properties of the atoms in
them.
They are also called pure semiconductors.
Some electrons in intrinsic semiconductors are held less tightly than other.
2. Extrinsic semiconductors:
These are semiconductors which are obtained by doping.
Doping is the process of adding minute quantities of substances called impurities.
The importance of doping is that conductivity of the semiconductor is greatly improved.
The doping process leads to two types of extrinsic semiconductors which are:
o N-types semiconductor.
o P-type semiconductor.
a. N-type semiconductor:
This is a type of extrinsic semiconductors whose conductivity is due to presence of an
extra electron.
140
It is done by doping a semiconductor with 4 electrons in the valence shell by a penta-
valent atom (an atom with 5 electrons in the valence shell).
If for example, silicon is doped with phosphorous one of the electrons in the
phosphorous is not included in the lattice bonding. This results in more free electrons
in the semiconductor which in turn increases conductivities several times.
Since the majority of the charge carries are electrons with a negative charge then these
semiconductors are called n-type semiconductors (n means negative).
b. P-type semiconductor:
This is due to absence of electrons which implies presence of a positive hole.
It is obtained when an atom with four electrons in the valence shell is doped with a tri-
valent atom (an atom with 3 electrons in the valence shell).
For example, when silicon is doped with boron there will be 7 electrons in the lattice
bonding and this missing of an electron results in positive holes.
Since the majority of charge carries are positive holes they are called p-type
semiconductors (p means positive).
Impurities which release free electrons are called donors (e.g. arsenic, antimony and
phosphorous) and those that receive the free electrons are called acceptors (e.g.
indium, gallium and boron).
USES OF SEMICONDUCTORS:
Semiconductors have got wide application in the electronics industry.
They are used in designing of diodes, transistors, thermistors and also in integrated
circuits.
DIODES:
A diode is a two terminal, one way device which allows current to flow in one direction.
The wire nearest the band is a called cathode and the other is called anode.
The diode conducts when the anode is connected to the anode side of a cell or battery.
It is then said tobe forward biased. The resistance is small and current flows in the direction
of the arrow of its symbol.
When the diode is reversed is said to be reverse biased and there is a large resistance
and hence no conduction.
In diodes, a singe crystal of silicon or germanium is doped in such a way that one half of
it is p-type and the other is n-type.
141
USES OFDIODES:
a. As a current rectifier.
It allows current to flow in one direction. This process is called rectification.
Thus, rectification is the conversion of alternating current into a direct current.
Use of a single diode results into a half wave rectification where by the original current
value reduces to half.
Use of a net-work of diodes called diode bridge leads to a full wave rectification.
b. As a resistor:
Diodes can also be used as resistors. They limit current flow. Flow of current depends on
how the diode has been connected to terminals of the voltage source which determines
whether the diode is forward biased or reverse biased.
USES OF TRANSISTORS:
A transistor is a semiconductor device used to amplify and switchelectronic signals and power.
It has three connections; base b, emitter, e and collector, c.
Most transistors are silicon n-p-n types. In this one a single crystal of a semiconducting
material is doped in such a way that a piece of p-type is sand-witched between two
pieces of n-type material.
The base-emitter junction is normally forward biased and the base collector is reverse
biased when the transistor is in use.
The following is the circuit symbol of a transistor:
In a transistor there are two current paths
I. base-emitter path
II. Collector – emitter path.
142
There are two types of transistors and these are:
a. n-p-n type
b. p-n-p type.
The transistors are normally used in the following:
a. as switch:
E.g.:
I. Light sensitive switch
II. Time-delay switch
III. Temperature dependent switch
This is the most important use of transistors. A transistor links circuits connected to
each other so that current in one path controls that in the other path. I.e. they can be
used as automatic switches such as relay.
Advantages of transistors:
Over other electrically operated switches:
I. They are small
II. They are cheap
III. They are reliable
IV. They have no moving parts
V. Have indefinite life span once in a well-designed circuit
VI. Can switch on and off million times in second.
b. As current amplifier.
An amplifier is a device that magnifies the input.
Electronic amplifiers magnify small input signals and produce large signals. The input
might be current or voltage.
In case of a forward biased (n-p-n) transistor, I
E
= I
B
+ I
C
.
c. As voltage amplifiers:
Amplifies are normally used to provide voltage amplification.
THERMISTORS:
A thermistor a semi-conductor device which changes its resistance as the temperature
changes
As temperature increases the resistance of a thermistor decreases
This property of the thermistor is used in sensing temperature changes as electronic
thermistor in a refrigerator.
The symbol for a thermistor is
143
HOW THE TOPICS (ELECTRICITY AND MAGNETISM 1 & 2) HAVE BEEN FEATURED IN THE LAST 4
YEARS OF MANEB EXAMINATIONS
(Find some suggested answers in italics).
2015 No 2
a. Figure below shows diagram of an electric circuit.
(i) Name the part labeled X Answer: resistor
(ii) State two uses of the device labeled Y Answer: as switch and as amplifier
(iii) Explain the importance of the small base voltage in the circuit
Answer: it changes the behavior of the transistor (device Y). It removes the
blocking effect hence switching it on
(iv) Which device could indicate that the current is flowing in the circuit?
Answer: the bulb
(v) Explain how the device mentioned in 2a(iv) works
Answer: when the small base voltage switches on the transistor current flows
and the bulb gives light and when there is no any current in the base of device
y, the transistor is off and current no longer flows hence the bulb gives no light.
b. A transformer steps down voltage from 240V to 120V to operate a hair drier.
Calculate the current in the primary coil if the current flowing in the hair drier
is 10A
The question did not give the efficiency of the transformer. Assuming it is 100%
then
AI
IIVIV
p
psspp
5
240
1200
10120240
144
c. Explain how leakage of magnetic field lines reduces the efficiency of a
transformer
Answer: the leakage of field lines affects the voltage in the secondary coil and make it NOT
proportional to the voltage in the primary coil
2014 No 6
a.
(i) What are semiconductors?
Answer: These are materials that conduct electricity partially
(ii) Name any two elements that can be used to dope silicon
Answer: Boron, phosphorous
b. Why does electrical resistance of metals increase as the temperature of the metal is
increased?
Answer: the increase in temperature causes huge vibration and hitting of free
electrons with stationary positive nuclei. This slows down speed/flow of electrons
c. Figure below shows a resistor whose resistance is indicated in standard notation
Work out its resistance in ohms
2013.
6c calculate the current flowing in the primary coil of a 100% efficient transformer that steps
down voltage from 240V to 20V if current flowing in the secondary coil is 10A
7.
a. Mention any two factors that affect the efficiency of a transformer
Production of eddy currents
Leakage of field lined / flux leakage
Resistance of wire windings
%203200
%202.3
or
K
AI
I
IVIV
p
p
sspp
83.0
240
1020
1020240
145
b. With the aid of a well labeled diagram, explain how a step down transformer works
A step down transformer has more wire turns and higher voltage in the primary
coil than in the secondary coil. It has a core made of iron for easy magnetization.
When a voltage is applied to the primary coil, it magnetizes the iron core, which
induces voltage in the other coil (called secondary coil) by cutting through the wire
coils.
2012 2aState two ways of inducing electromotive force (emf)
I. By moving wires coils up and down at right angles with a magnetic field
II. By moving a bar magnet into and out of a wire coils.
b Mention any one application of electromagnet
functioning of an electric bell, designing of magnetic relay switches etc
c Figure below shows an electric circuit.
(i) What is the function of the part labeled Z?
To break the circuit when too much current flows
(ii) State any two advantages of the circuit above
I. Components work independently
II. If a component is faulty, the other components still work
III. In case of bulbs, addition of more bulbs has no effect on brightness of the others
(iii) Which bulb would use more current if both switches were closed? The one rated
200W because in a second it will be converting 200J of electrical energy to heat plus
light. (Has highest power which is directly proportional to current)
146
d Calculate the power dissipated in an electric heater in which 4A of current flows when
connected to a 230V supply
8e. Explain how a piece of steel could be magnetized by single touch stroking method.
This could involve moving one bar magnet repeatedly, in the same direction
above the steel.
10.OSCILLATION AND WAVES
Topic Objectives
By the end of your interaction with this topic you should be able to:
explain the term oscillation
describe the characteristics of an oscillating system
describe a wave
Explain characteristics of a wave
distinguish a transverse wave from a longitudinal wave
describe wave properties
use the wave equation in problem solving
distinguish between a convex lens and a diverging lens
determine the focal length of a converging lens
explain image formation by a converging lens
draw ray diagrams
apply the lens formula in problem solving
discuss functions of parts of a camera
describe how a camera works
WVAxVIP 9202304
147
draw comparison between a camera and an eye
discuss parts of a projector and their functions
describe how a projector works
Oscillation:
it refers to a periodic motion in which a body retraces its path at equal time intervals
It is also defined as a to and fro movement of an object.
A periodic motion is observed in a simple pendulum which consists of a mass hung at
the end of a string. refer to the fig below:
Characteristics of an Oscillating System
Characteristics of an oscillating system defers from properties of the oscillating system hence in
this topic the words ‘characteristics’ and ‘ properties’ should not be used interchangeably. The
list of characteristics of an oscillating system includes:
Amplitude
period
frequency
wavelength
speed
1. Amplitude, A
This is the maximum displacement reached by an oscillating system from its resting
position.
the resting position is also known as an equilibrium position
The amplitude can be given in meters, centimeters or millimeters.
2. Period, T
This is the time taken for an oscillating system to complete one oscillation (cycle).
It can also be defined as the time taken for one complete vibration
The SI unit of period, T is the second (s)
3. Frequency, f
This is the number of cycles of an oscillating system per second.
148
the SI unit of frequency is the hertz ( Hz)
The period, T and the frequency, f are just reciprocals of each other. i.e.
T
f
1
f
T
1
.
For example, the escom’s electricity is a.c and has a frequency of 50Hz. calculate its
period. Solution :
f
T
1
.
onds
s
T sec02.0
/50
1
. Note
ondcycleHz sec/11
4. wavelength, λ
This is the distance between two consecutive crests or troughs.
It is also defined as the distance between two consecutive particles in phase.
A crest is the highest point reached by a vibrating particle while a trough is the
lowest point reached by the same.
5. speed, v
This is the distance moved by a crest or a trough or any point on the wave in one
second. It’s normally given in meters per second.
The figure below shows parts of a wave
Crest wavelength Crest
A
Distance
Trough
Displacement against distance. A is the Amplitude
Crest Period Crest
Time (s)
Displacement against time
WAVE EQUATION
If a particle spends t seconds in moving from one point to the other then the speed, v will be as
follows:
149
time
cedis
speed
tan
Now consider the distance from one crest too the next which is the
wavelength and time taken from one crest to next which is the period then Speed=
wavelength/period. I.e.
T
v
Since
f
T
1
. Then
f
v
1
fv
Where v = speed in m/s
λ = wavelength in m
f = frequency in Hz
T = period in seconds.
Example 1
a.
Distance 47.5 m Distance (m)
I. Calculate the wavelength of the wave represented above.
II. If the period of the wave above is 10 seconds, what is the speed of the wave?
Solution:
I. there are
5.9
2
1
9
complete waves hence wavelength =
m
m
5
5.9
5.47
II.
fv
. Need to find the frequency.
Hz
sT
f 1.0
10
11
thus
smmHzxv /5.051.0
Example 2
A wave travels a distance of 50 cm in 5 seconds and the distance between two successive
troughs is 2.0 m
Calculate
a. Velocity
b. Frequency of the wave
Solution
150
a.
sm
s
m
time
cedis
v /1.0
5
5.0tan
b.
fv
Hz
m
smv
f 05.0
0.2
/1.0
Example 3.
Distance( m)
0 2 4 6 8 10 12
Displacement against distance in meters
Work out the following:
a. wavelength
b. Speed of the wave if its frequency is 2Hz.
Solution
a. The wavelength is the distance from one crest to the next hence the wavelength is 4m.
b.
smmHzxfv /842
Example 4.
time, s
0 1 2 3 4 5 6
Displacement against time in seconds
a. find the frequency of the wave
b. If the wave is traveling at 5 m/s, what is its wavelength?
Solution
a. Need to find the period. It is the time taken for a wave to complete one cycle. (From a crest to
the next) then Period = 2 seconds.
Therefore
Hz
sT
f 5.0
2
11
b.
m
m
sm
f
v
10
5.0
/5
151
o the energy transmitted by waves depends on frequency and amplitude
o in the absence of friction, oscillating systems obey the principle of conservation
of mechanical energy ( PE + KE)
o However, all oscillating systems lose energy and their vibrations dieaway and are
called damped oscillations.
o Examples of oscillating systems include: a simple pendulum, a spiral spring, a
ruler (cantilever), a watch, a swinging rope etc.
o A wave, like all other oscillations have characteristics such as amplitude, period,
frequency, wavelength and velocity (speed).
Factors That Affect Frequency of Some Oscillating Systems.
a. simple pendulum
I. The length of the pendulum
The shorter the length the higher the frequency and vice-versa
II. material from which the pendulum is made
III. amplitude
The smaller the amplitude the higher the frequency and vice-versa
Note: the mass of a bob does not affect frequency. When doing experiments on these,
only one factor should be changed the rest have to be controlled.
b. Spiral spring
I. Mass on spring
The higher the mass hung at the end the smaller the frequency.
II. Material from which the spring is made.
Some materials make a spring stiffer or weaker and this affect how they
oscillate.
Note: Length of spring and amplitude changes do not affect the frequency of the oscillating
system.
c. Cantilever
I. mass on end
The smaller the mass the higher the frequency and vice-versa
II. the material from which the cantilever is made
III. length of the cantilever
The shorter the length the higher the frequency and vice-versa
Amplitude changes do not affect the frequency of a vibrating cantilever
CLASSIFICATION OF WAVES
The classification depends on the criteria used. if one considers media for transmission
then the waves are classified as
a. mechanical waves
b. Electromagnetic waves.
a. mechanical waves:
152
These are waves that require a medium for transmission. Examples of media are water,
air etc.
Examples of waves in this category are: water wave, sound waves, waves in a stretched
string, seismic waves which travel through earth’s crust.
When a mechanical wave travels from some point A to some point B, it is because a
disturbance of some kind at A has caused particle to become displaced.
The displaced particle drugs its neighbor with it so that it too gets displaced and has a
similar effect on the next particle. This continues until the disturbance reaches point B.
b. Electromagnetic waves:
These are waves which do not require a medium for transmission.
It is as a result of vibration of a particle in a magnetic and electric fields.
Examples of electromagnetic waves include radio waves, x-rays, microwaves, infra-red,
ultraviolet, gamma rays, light waves etc.
Electromagnetic spectrum:
This is a list of electromagnetic waves arranged in accordance to their frequencies and
wavelength.
The table below shows examples of electromagnetic wave, their frequencies and uses.
DIVISION
FREQUENCY
USES
Radio waves
3x10
4
to 3x10
9
Radio communication
microwaves
3x10
10
Radar, cooking
Infra-red
10
11
to 10
14
Heating, night sights
Visible light rays
5 x 10
14
Stimulates the retina, initiates photosynthesis
Ultraviolet (UV)
10
15
Promote chemical reaction. used in photography
X-rays
3 x 10
18
Used in medicine to
I.locate borne fractures
II.destroy cancer cells
III.locate imperfections in welding and casting
Gamma rays
3 x 10
19
To sterilize medical equipment & destroy cancer cells
If one looks at how the waves are propagated then we would have the following types
of waves:
Types of waves
a. Transverse
b. Longitudinal waves
a. transverse waves:
These are waves in which the displacements are perpendicular to the direction
of travel or propagation.
Examples are water waves, light waves and all the electromagnetic waves
Their general shape is as follows:
153
b. Longitudinal Waves
these are waves in which the displacements are parallel to the direction of the
propagation
examples are: Sound waves, Tsunami waves, waves in a slink, internal water
waves, spring oscillations
Both types of waves can best be demonstrated by a long steel spring called a
slinky.
Similarities between transverse and longitudinal waves:
1. both waves can be represented by sine waves
2. both can be reflected
3. they can be refracted
4. both get involved in interference
5. both can be diffracted
Differences between transverse and longitudinal waves:
1. Transverse waves can be polarized (confined on the same plane) while longitudinal
waves cannot be polarized.
2. In transverse waves particles vibrate perpendicularly to the wave direction while in
longitudinal waves particles vibrate parallel to the wave propagation.
3. Transverse waves include all the electromagnetic waves while no any electromagnetic
wave is longitudinal.
PROPERTIES OF WAVES:
a. Reflection of Waves:
This is the bouncing back of waves when it meets a barrier
When a wave gets reflected the angle of incidence is equal to the angle of refection.
b. Refraction of Waves:
154
This is the bending (deviation) of waves when they cross boundary between two different media.
During refraction the wavelength and speed of the waves change so too with the direction.
Frequency remains constant
Water waves travel more slowly in shallow waters and as a result they are refracted towards the
normal as they enter shallow region. This is so because since
fv
and f remains constant while
λ decreases leading to decrease in velocity in shallow waters.
The figure illustrates bending of water waves when they move from deep regions to
shallow regions.
c. interference or superposition of waves
It is the addition ( superposition) of two or more waves which results into a new wave
pattern
If a crest meets a crest or a trough meets a trough then there is a constructive interference
and the wave produced has large amplitude.
In other words constructive interference occurs when the waves are in a phase.
If the waves are out of phase the destructive interference occurs.
Destructive interference occurs when a crest meets a trough.
water waves barrier shallow region
deep
region
155
156
d. Diffraction
This is the bending (deviation) of waves in a single media when it passes through a narrow
aperture.
During diffraction there is no change in wavelength or speed but amplitude increases.
The size of the slit (gap) affects how a wave spreads out.
If the slit is narrow the waves curve more and there is a linear propagation if the gap is
made wider.
Diffraction
157
If there are two gaps then two wave properties will be observed and these are diffraction and
interference. Refer to the figure below.
LIGHT AS AN EXAMPLE OF ELECTROMAGNETIC WAVES
A light is a transverse wave
It travels at the speed of 3 x 10
8
m/s.
LENSES
A lens is a grass or a plastic prism with curved surfaces.
It can be converging or diverging
a. Converging lens.
they cause light rays to come to a similar position
They are also called convex lens.
They are thicker on the middle and narrow on the sides.
b. diverging lens
They cause light rays to spread out.
They are also known as concave lens.
They are narrow on the middle and wider on the edges.
Convex lens
(converging lens)
Concave (lens)
(diverging lens)
Their Focal length is positive
Their focal length is negative
After refraction, Light rays
meet(appear to meet) at one
point
After refraction, light rays move
away from one another
The image they form can be
real, virtual, magnified or
diminished depending on the
object distance
They always form an image
which is virtual and diminished
Used to correct long sight
Used to correct short sight
158
Important terms:
1. PRINCIPAL FOCUS (FOCAL POINT), F of a converging lens is that point on the principal
axis of the lens to which light rays parallel to it converge after refracting at the surface
of the lens.
2. PRINCIPAL AXIS of the lens (converging or diverging) is the line which passes through
the centre of the lens surface.
3. FOCAL PLANE is an imaginary plane through F which is perpendicular to the principal
axis.
4. REAL IMAGE is one through which rays of light actually meet.
A real image
Can be cast onto a screen.
Is always inverted.
5. VIRTUAL IMAGE is the one from which rays of light only appear to have come from.
A virtual image
Cannot be cast onto a screen.
Is always upright.
6. FOCAL LENGTH (f) is the distance from the lens to the principal focus.
7. IMAGE DISTANCE (V) is the distance from the image to the lens.
159
8. OBJECT DISTANCE (U) is the distance from the lens to the object.
9. OPTICAL CENTRE is the middle part of a lens.
RAY DIAGRAM.
Important ray path:
All the rays of light that leave a single point on an object meet at the same image point.
A ray, through the optical centre of the lens is un-deviated.
A ray, parallel to the principal axis, passes through ( appear to leave or come from F)
A ray, through (or heading towards) F, emerges parallel to the principal axis.
1) Object beyond 2F, image between F and 2F, real, inverted and diminished.
2) Object at 2F, image at 2F, real, inverted and of the same size as the object.
3) Object between 2F and F, image is beyond 2F, real, inverted and magnified.
160
4) Object at F, no image formed since light rays meet at infinity.
5) Object between F and O, image virtual, upright and magnified and is on the same side as
the object.
161
THE LENS FORMULAR
This is given by
fvu
111
where u = object distance, v = image distance and f = focal
length of the lens.
The focal length of a converging lens are positive, those of a diverging lens are negative.
Distance from the lens to real images is positive whereas the distance from the lens to the
virtual image is negative.
MAGNIFICATION
Magnification of an object is given by
htobjectheig
timageheigh
ceobjectdisu
ceimgedisv
m
)tan(
)tan(
Formation of an image by a diverging lens, for all positions of the objects, the image is
virtual, erect and smaller than the object and is situated between the object and the lens.
Examples
1. An object 9cm high is placed 24cm in front of a converging lens and real image is formed
8cm away from the lens find
a. Size of the image
b. The focal length of the lens
a.
cmh
h
htobjectheig
timageheigh
ceobjectdisu
ceimgedisv
3
24
98
924
8
)tan(
)tan(
2
2
b.
cmf
f
fvu
6
4
24
24
4
24
311
8
1
24
1
111
EXPERIMENTAL DETERMINATION OF FOCAL LENGTH OF A CONVERGING LENS:
1) Distance Object Method.
o Position the lens so that it produces a sharply focused image of a distant object (e.g. a
window or a screen.
o The distance between the lens and the screen or the window is roughly taken to be the
focal length of the lens.
2) Light Experiments.
o By using an illuminated object and a screen, obtain a number of values of U and the
corresponding values of V.
o The graph of
U
1
against
V
1
is plotted and each intercept of the graph is equal to
f
1
and
hence f is found.
162
E.g. consider the values given in the table below:
U (cm)
V (cm)
U
1
(cm
-1
)
V
1
(cm
-1
)
12.5
49.9
0.080
0.020
15.0
29.9
0.069
0.033
20.0
20.0
0.050
0.050
25.0
16.7
0.040
0.060
30.0
14.7
0.033
0.068
35.0
13.7
0.029
0.073
40.0
13.0
0.025
0.077
o When corresponding values of u and v are collected as In method 2, the graph of ( U+V) against U
can also be used to find the focal length of a lens.
U+V (cm)
u+v = 4f
u=2f U (cm)
Example
163
In an experiment to determine focal length of a convex lens, the following data was obtained
Object distance (cm)
U
Image distance (cm)
V
(U+V) cm
12
58
15
30
20
20
30
15
40
13.3
50
12.5
(i) Complete the table under column of (U+V)
ii. Using the graph paper provided, Plot a graph of (U+V) against U
Object distance (cm)
U
Image distance (cm)
V
(U+V) cm
12
58
70
15
30
45
20
20
40
30
15
45
40
13.3
53.3
50
12.5
62.5
164
Focal length
cmff
or
cmff
10
4
40
404
10
2
20
202
o Other methods are:
3) Plane mirror method ( self-conjugate foci)
4) Displacement method.
OPTICAL INSTRUMENTS:
These are instruments which depend on light for their functioning.
Examples are: camera, projector, binoculars, and microscope.
THE CAMERA:
This is a device whose main function is to provide a dark place to hold a film.
Camera is a light-tight box in which a convex lens forms a real image on a film.
some parts of a camera
I. Iris diaphragm
This works as the eyes-iris does. Normally the diaphragm is closed. When a picture is to be taken
in dark area, the diaphragm opens up to provide a large hole called aperture. The aperture is the
pupil of the eye.
165
II. The shutter
This controls the light entering the camera by opening and closing the aperture at different
speeds.
III. Photographic film
This is where the images are focused just like on the retina of the eye.
IV. The lens
This refracts light.
It brings the images to focus on the film no matter what the distance of the object is, or whether
it is in the center of the scene or towards the edge.
Shutter speed control
Lens Film
Shutter
Diaphragm
Picture of a camera and how the lens refracts light
SIMILARITIES BETWEEN A CAMERA AND AN EYE:
CAMERA
EYE
Has a light sensitive film
Has a light sensitive retina
Has a converging lens
Has a converging lens
Inside surface is black
Inside surface is back
Light controlled by shutter diaphragm
Light controlled by the iris of the eye
Lens forms real, diminished and inverted
image
Lens forms real, diminished and inverted image
DIFFERENCES BETWEEN A CAMERA AND AN EYE:
CAMERA
EYE
Focal length of lens is fixed
Focal length of the lens can be altered
Closed except when taking a picture
Pupil normally open
Image distance changes
Image distance fixed
EYE DEFECTS:
Accommodation:
It is the automatic adjustment in focal length of the natural lens of the eye.
There are two types of eye defects which are: short sight and long sight.
166
Short Sight (myopia):
o Ability to see near objects clearly.
o The image of a distant object is focused in font of a retina due to long eye balls.
o It is corrected by using spectacles with diverging lens.
Long sight (hypermetropia or hyperopia)
o Ability to see distant object but failing to see near ones.
o It happens due to short eye balls. The image is focused behind retina.
o A spectacle with convex (converging) lens corrects the problem.
o Note that if a person holds a book at arms length then he/she is long-sighted.
167
THE PROJECTOR:
This is a device which is used to form a magnified image of a small object (such as a
photographic slide) on a screen placed at several meters away from the lens.
It has two lenses and a concave mirror plus a bulb to illuminate the film.
Picture of a projector
Concave mirror Slide
Image
Light source Condenser lens Projector lens
Screen
Concave mirror
This reflects light from the bulb onto the film (slide).
Condenser Lens (Condensing Lens)
This is used to
converge light and
Concentrates as much light as possible on the film so that it is very bright and evenly lit.
Projection Lens:
It is used to produce considerable, focused, magnified, real and inverted images of the film.
The distance from the lens to the film must be between F and 2F.( Just outside F)
The projector lens has a longer focal length than the condenser lens.
Upright images are produced by placing the film (object) upside down.
To increase magnification one can do the following:
Increase image distance, V and decrease the object distance, U. This can be achieved by
Moving the projector lens nearer to the film or object.
Moving the film/object nearer to Projector lens.
Moving the screen away from the projector lens.
168
HOW THE TOPIC HAS BEEN FEATURED IN THE LAST 4 YEARS OF MANEB EXAMINATIONS
(Find some suggested answers in italics).
2015 N0 4
a. Mention any two properties of waves
Answer: reflection, refraction, interference and diffraction
b. (i)State any two differences between convex and concave lenses
Convex lens
Concave lens
Thick in the middle than edges
Light rays are bent inwards
Narrow in the middle
Light rays are bent outwards
(ii) Draw a ray diagram to show the position of an image 4cm high placed 10 cm from a
convex lens of focal length 5 cm. (use a scale of 1 cm to represent 2 cm )
(iii) Calculate the magnification of the image
Answer:
1
10
10
u
v
mg
c. Figure below shows two waves travelling in opposite directions
(i) Draw a diagram to show what happens when the two waves meet
169
(ii)Draw a diagram to show what happens when the two waves pass each other
(iii) What property of wave is demonstrated in 4c(i)?
Answer: Interference
2014 6
d. Mention any three characteristics of waves
Speed, Frequency, Wavelength, Amplitude, Period
e. Figure below is a diagram showing a wave
(i) What is the amplitude of the wave?
Answer: 0.2m
(ii) What type of wave is shown in the diagram?
Answer: Transverse wave
(iii) Calculate the frequency of the wave is its speed is 8m/s
8a explain how a slide projector works
It produces a real magnified image through functions of a small bulb which provides light;
condenser lens which converges and concentrates light onto a film. It has a projector lens
which refracts light and focuses it on the screen where the image is formed. The slide (film) is
put up-side down in order to obtain an upright image and furthermore, the slide is put
between F and 2F in order to get a magnified image.
2013
a. Mention two differences between convex and concave lenses
Hz
m
smv
f
fv
4
2
/8
170
Convex lens
(converging lens)
Concave (lens)
(diverging lens)
Their Focal length is positive
Their focal length is negative
After refraction, Light rays
meet(appear to meet) at one
point
After refraction, light rays move
away from one another
The image they form can be
real, virtual, magnified or
diminished depending on the
object distance
They always form an image
which is virtual and diminished
Used to correct long sight
Used to correct short sight
b. Figure below shows part of ray diagram
(i) Complete the ray diagram to locate the position of the image
The answer below is not to scale. Exact position would be obtained by accurate
drawing.
(ii) State the nature of the image formed
Image would be virtual, upright and magnified
c. Mention any two similarities between the human eye and a camera
CAMERA
EYE
Has a light sensitive film
Has a light sensitive retina
Has a converging lens
Has a converging lens
Inside surface is black
Inside surface is back
Light controlled by shutter diaphragm
Light controlled by the iris of the eye
Lens forms real, diminished and inverted
image
Lens forms real, diminished and inverted image
171
d. Why are radio waves classified as transverse waves?
Answer: because particles vibrate perpendicularly to the direction of the wave
2012 4
a. Define “oscillation”
It is defined as a to and fro movement of an object
b. (i) What type of a wave is produced by a vibrating string? Transverse wave
(iii) Calculate the frequency of a wave with a wavelength of 2m and speed of 6m/s
c. State the difference between “interference” and “diffraction” of waves. Diffraction
is the bending (deviation) of wave in a single media when it passes through a narrow
gap while interference is the addition ( superposition) of two or more waves which
results into a new wave pattern
8.
a. With the aid of a well labeled diagram, explain how destructive interference in water
waves occurs.
Water waves, which are circular in nature spread out from a source and overlap
(interference). If a crest of one wave meets trough of another wave then the
destructive interference occurs.
b. Explain why waves refract when travelling from one medium to another. It’s due to a
decrease in speed. The wavelength reduces but the frequency remains constant.
Hz
m
smv
ffv 3
2
/6
172
11. NUCLEAR PHYSICS
By the end of this chapter learners should be able to:
Name constituent particles of the atomic nuclei
Express composition of a particular nucleus in standard notation
Describe isotopes
Explain what is meant by radioactivity
State characteristics of radioactive substances
Describe alpha and beta particles and gamma rays
Describe detection of radiation
Explain the meaning of radioactive decay
Distinguish between natural and induced radioactivity
Explain and use the idea of half-life
Describe safety precautions during handling and storing of radioactive substances
Distinguish between nuclear fission and nuclear fusion.
Describe the use of radioactivity
STRUCTURE OF AN ATOM
An atom is a basic unit of matter which contains other three sub-atomic particles. These are
protons, electrons and neutrons. The table 1 below summarizes their location, mass and type of
charge.
PARTICLE
SYMBOL
LOCATION
MASS
CHARGE
Electron
E
Shell
0
negative
Neutron
N
nucleus
1amu
zero
proton
P
nucleus
1amu
positive
Note: the mass of an electron is almost zero (negligible) as compared to the mass of the proton.
amu means atomic mass unit. The shell is also called an energy level.
STANDARD NOTATION
The standard way of expressing composition of an atom involves showing mass number, atomic
number and element symbol. Refer to the following:
X
A
Z
where A is the atomic mass or mass number
Z is the atomic number. It gives the position of an element in the periodic table and it is equal
to the number of protons.
Mass number, A = Atomic No. + No. of neutrons
173
Therefore, No. of neutrons = Atomic mass – Atomic number
i.e. No. of neutrons = A – Z. Refer to the following example:
Given the following:
Al
27
13
. Determine
a. The mass number
b. Atomic number
c. Number of neutrons
d. Number of electrons.
Solution:
a. The mass number = A = 27
b. Atomic number = Z = 13
c. No. of neutrons = A – Z = 27 – 13 = 14
d. No. of electrons = No. of protons = Atomic number = 13
ISOTOPES
These are atoms of the same element having the same atomic numbers but different mass
numbers. The difference in mass numbers is due to difference in number of neutrons.
For example the nuclides
O
16
8
,
O
17
8
and
O
18
8
are isotopes of oxygen while
Na
23
11
and
Na
24
11
are isotopes of sodium.
Isotopes have similar chemical properties due to the same number of electrons.
Isotopes can be stable or unstable
Stable isotopes are isotopes in which the number of both neutrons and protons are
about equal and are within tolerable levels or comparable ratios e.g.
O
16
8
,
O
17
8
and
O
18
8
Unstable isotopes are those in which there are two many neutrons for the protons and
vice- versa. E.g.
U
238
92
,
Ra
226
88
and
Sr
90
38
Other examples of isotopes include those of hydrogen which are: Hydrogen
H
1
1
also
called Protium, Deuterium
H
2
1
or
D
2
1
and Tritium
H
3
1
or
T
3
1
RADIOACTIVITY OR RADIOACTIVE DECAY
This is the spontaneous disintegration of unstable nucleus to acquire a more stable
state.
Radioactivity happens in order to equalize or normalize its proton neutron ratio.
A radioactive substance is the one that breaks in order to gain a stable form.
Examples of radioactive substances are: Uranium
U
239
92
, Radium
Ra
226
88
, Polonium
Po
210
84
During radioactive decay, energy is released. In addition, several particles are emitted
which include alpha, beta and gamma rays.
Nuclear radiation is the energy released when radioactivity occurs.
Take note that a particle can still be undergoing decay but without releasing energy.
174
TYPES OF DECAY PARTICLES
ALPHA (α- PARTICLE)
It is a radioactive emission identical to a helium nucleus
The following is its symbol:
4
2
or
He
4
2
Characteristics of alpha radiation
This consists of two protons and two neutrons
As already mentioned, it is identical to a helium nucleus
When a nucleus undergoes α- decay it loses four nucleons, two of which are protons.
Therefore
I. Its mass number (A) decreases by 4
II. Its atomic number (Z) decreases by 2.
i.e.
HeYX
A
Z
A
Z
4
2
4
2
(Parent) (Daughter) α – particle (decay product)
Example: consider the decay of Uranium- 238
HeYU
4
2
234
90
238
92
Properties of Alpha particles
1. It is the most highly ionizing particle. I.e. it knocks off electrons in an atom and
makes it charged. This is attributed to the high charge they have ( +2)
2. It can penetrate about 5 cm of air because of its relatively large mass.
3. It can be deflected by electric and magnetic fields. (It bends in these fields). This is so
because of its positive charge and any charge creates a magnetic and an electric
field.
4. It is positively charged( has a charge of +2)
5. Move at a low speed as compared to the speed of light (due to its large mass).
Other examples of α-decay include the following:
HePoRn
4
2
218
84
222
86
HePbPo
4
2
214
82
218
84
BETA (β- PARTILCE)
It is a radioactive emission identical with an electron.
Its symbol is as follows:
0
1
or
e
0
1
Characteristics of beta radiation
These are simply high energy and fast moving electrons emitted by any given species of
nucleus. The nucleus has too many particles to be stable.
175
In the emission of beta particles one of the element’s neutrons transforms to a proton
and an electron. The proton remains in the nucleus.
Beta rays are streams of high energy electrons.
When a nucleus undergoes β- decay
I. Its mass number (A) does not change
II. It atomic number (Z) increases by 1
In general,
eYX
A
Z
A
Z
0
11
Example:
I.
ePSi
0
1
27
15
27
14
ii.
eBiPb
0
1
214
83
214
82
Properties of Alpha particles
1. Produces much less ionization than alpha particle. It’s due to its relative low charge.
2. It can penetrate 500cm of air as well as 2mm of aluminum since it is lighter as compared
to alpha particle.
3. They are negatively charged
4. They are easily deflected by both electric and magnetic effects (Due to the negative
charge they have).
5. They move at about 0.9 x the speed of light ( 3x10
8
m/s)
GAMMA (
-RAYS)
It is a radioactive emission of very short wavelength of electromagnetic radiation.
Its symbol is
0
0
or
Properties of gamma rays
1. They produce very little and weak ionization effect because their charge is zero.
2. They are very penetrating. Thus, they are able to penetrate through about 4cm of lead
and their intensity is reduced by thick concrete. This is due to the fact that they are very
light and move at a high speed, a speed of light.
3. They are not deflected by magnetic and electric fields because they do not carry any
charge.
4. They have no mass and charge
5. Have a speed as that of light
COMPARING THE LEVEL OF PERNETRATION:
Figure below shows relative penetration power among the three types of radiation.
Invisible
Nuclear β
Source α
Paper aluminum lead
176
DESCRIBING DEFLECTION IN AN ELECTRIC FIELD
Figure below shows radiation passing through an electric field.
Radiation source
Positive plate
Negative
Plate α
β
The alpha particles get attracted to the negative plate since they are positively charged and
unlike charges attract each other. The same applies to the beta particles which are attracted
towards the positive plate since they have a negative charge. Gamma rays are not affected in
the field since they do not possess any charge to warrant either repulsion or attraction. Note
that beta particles bend more than alpha particles because beta particles have a smaller, if not
a negligible mass as compared to that of alpha (helium) particles.
TYPES OF RADIOACTIVTY
Radioactivity can either be natural or artificial (induced)
a. Natural radioactivity
This happens without human interference and involves mostly nuclides whose atomic
numbers are more than that of lead
Pb
207
82
Examples of nuclides which can undergo natural radioactivity are uranium-238, thorium-
234, radium-226 etc.
b. Artificial or induced radioactivity
This is done by exposing a stable naturally occurring nuclide to different atomic particles
inside a nuclear reactor. The different particles used in this bombardment are: neutrons,
protons, hydrogen isotopes etc.
Take note that the induced radioactivity leads to production of energy and different
radioisotopes.
Next part gives examples of how induced radioactivity can be achieved;
1. Neutron bombardment
- - -
- -
- -
- -
- -
- -
- -
- -
- -
++++
++++
++++
++++
++++
++++
++++
++++
++++
177
When a stable non-radioactive uranium 238 is bombarded by a neutron particle a
radioactive uranium-239 is obtained
UnU
239
92
1
0
238
92
2. Alpha bombardment
nNaF
1
0
22
11
4
2
19
9
HON
1
1
17
8
4
2
14
7
A proton or neutron gets emitted in the process.
Description of a proton bombardment is also done as above.
SOURCES OF RADIATION:
Sources of radiation can be categorized into two namely:
a. Natural sources
b. Artificial (human) sources
a. Natural sources of radiation
Three major sources of naturally occurring radiation are:
Cosmic radiation
This comes from the sun and outer space and consists of positively charged particles as
well as gamma radiation.
Terrestrial radiation
These are natural sources of radiation in the ground rocks, building materials and
drinking water supplies. Examples under this category include: Uranium, radium,
thorium and radon gas. The radon gas is as a result of decay of natural uranium in the
soil. The radon which emits alpha radiation rises from soil under the houses and can
build up in homes.
Internal sources.
Our bodies also contain natural radio nuclides. Potassium 40 is one example.
b. Artificial (Human) sources
The following information briefly describes example of human made sources of radiation.
Medical radiation sources
Examples include x-rays which have some similar properties to gamma rays.
consumer products
Examples include TV’s, older luminous dial watches, some smoke detectors etc.
Atmospheric testing of nuclear weapons.
This leads to residues of radioactive substances as a result of testing nuclear weapons
in the atmosphere and water bodies.
DETECTING RADIATION
178
There are a number of instruments used to detect radiation .this part describes some of the
widely used detectors.
I. By photographic plates.
- Once a photographic is exposed to radioactivity it gets fogged.
- The fogging of the photographic film is a measure of radioactivity to which they
have been exposed
- It was this phenomenon which lead to discovery of radioactivity by the scientist
Henri Becquerel in 1896.
II. By scintillation counter
- There are substances such as zinc sulphide which phosphoresce when affected
by radioactivity.
- As each particle from the radioactive source hits ‘the phosphor’ a flash of light is
emitted
- In scintillation counter, each flash of light gives rise to a pulse of current
- A digital counter records the pulse of current.
III. Spark counter
- It makes use of ejection (release) of electrons from the surface of metals when
light falls on them
- Sparks happen where radioactive particles ionize the air. ( this is normally the
case with the alpha and beta particles which have the ionizing power)
IV. Geiger- Muller Tube
- When a particle enters inside this tube it causes ionization of argon gas and a
flow of charge takes place
- The pulse of current caused by the flow of charge can be amplified and the pulse
is counted electronically by a scale or a rate meter.
V. Gold Leaf Electroscope
- A charged electroscope gets discharged when a radiation source (held in
forceps) is brought near the cap of the electroscope.
- Radiation knocks electrons out of surrounding air molecules leaving them as
positively charged ions. This is called ionization
179
- The positive air ions are attracted to the cap if the electroscope is negatively
charged and vice versa. This, in any case, causes the charge on the electroscope
to be neutralized.
Other radiation detectors are:
- Bubble chamber
- Wilson cloud chamber
- Solid state detectors
- Match counter
DECAY CURVES AND HALF-LIFE
The number of parent nuclei decreases exponentially with time and as a result it is known as
the exponential law of radioactive decay.
The radioactive decay is measured in activity per second or number of disintegrations per
second but the SI unit of radioactivity is called Becquerel, Bq.
The equation of decay is as follows:
N = N
0
e
-λt
……equation 1 where N is the amount remaining
N
0
is the original amount of nucleus
e is an exponent ( used when talking about natural logs)
t is the time taken for the decay
λ is the decay constant. OR the equation 2 which is
t
T
NN
2
1
0
Where T is the time for decay and t is half-life. N and N
0
are as above
HALF-LIFE:
This is the time taken for half of the nuclei to remain or time taken for half of the atoms in a
sample to decay. E.g. if before decay there were 60 grams of a radioactive sample, then after
180
half-life there would be 30 grams. After a further half-life there would be 15g then 7.5g and so
on.
Since after half-life the amount remaining (N) is half the original amount (N
0
) then it can be
stated as follows: at t= T
1/2
N= N
0
/2 then equation 1 becomes N
0
/2 =N
0
e
-λT
1/2
1/2 = e
-λT
1/2
ln ½ = -λT
1/2
T
1/2
= 0.6931/λ
Example 1.
Given that a radioactive particle of Bismuth-210 has a half-life of 5 days. If originally there is
80grams of Bismuth-210 determine the amount that would remain after
a. 5 day
b. 15 days
Solution:
Time in days
0
5
10
15
Amount remaining
80g
40g
20g
10g
a. 40g b. 10g
By use of formula; a. N=? N
0
= 80g t=5days, T= 5 days.
ggNN
t
T
40
2
1
80
2
1
5
5
0
b
ggggNN
t
T
10
8
1
80
2
1
80
2
1
80
2
1
3
5
15
0
OR
N=? N
0
= 80g t=5days. Need to find decay constant λ.
Using T
1/2
= 0.6931/λ thenλ = 0.6931/T
1/2
therefore, λ = 0.6931/5 = 0.13862
N = N
0
e
-λt
= 80 x e
-0.13862x5
= 40g
b. N = N
0
e
-λt
= 80 x e
-0.13862x15
= 10g
181
The table below shows some elements and their half-lives
ELEMENT
HALF-LIFE
Thorium
10
10
years
Radium-226
1602 years
Bismuth-210
5 days
Polonium-214
3 minutes
Iodine 131
8 days
Iodine 128
25 minutes
Carbon 14
5730 years
Boron 12
0.02 years
Radon 220
52 seconds
Example 2
If 6.25g of radioactive sample was found to remain after 208 seconds of radioactivity calculate
the half-life of the sample if the original activity was 100 grams.
N
0
= 100g, N = 6.25g T = 208 second, t = ?
.
2
1
2
1
0
0
t
T
t
T
N
N
NN
Thus
tt
208208
2
1
16
1
2
1
100
25.6
Therefore
.sec52
4
208
4
208
2
1
2
1
4
208
onds
s
t
t
t
In another way, N
0
= 100g, N = 6.25g t= 208 seconds T
1/2
= ?
Using the equation N = N
0
e
-λt
we have 6.25= 100x e
-λx 208
6.25/100 =
e
-λx 208
Ln (6.25/100) =
-
λx 208
λ = Ln (6.25/100)/(-208) = 0.0133
Now T
1/2
= 0.6931/λ = 0.6931/0.0133= 52 seconds
182
Graphical way of determining the half-life
The following table shows values for the decay of radioisotope iodine -131
Time (days)
0
8
16
24
Activity/ second
200
100
50
25
Plot a graph of activity against time. Use your graph to find the half-life of the iodine-131
clearly showing how you get it on the graph.
DANGERS, SAFETY PREACAUTIONS, AND STORAGE OF
RADIOACTIVE SUBSTANCES:
Dangers:
Beta and gamma rays can cause radiation burns( redness and sore on the skin)
It can also cause delayed effects like cancer and eye cataract
Low dosage over periods can cause fetal deformities, leukemia, bone cancer and other
forms of cancer.
Safety Precautions
Hold the radioactive substance using forceps.
Never hold a radioactive substance near one’s eyes
Put the substances in lead boxes when not in use
Avoid ingestion of the radioactive substance
183
Limit the amount of radiation you receive from x-rays and ultraviolet
Reduce time of contact with the radioactive substance
Never smoke near a radioactive source
Keep food away from places where radiation is used
Used sources must be dumped in concrete tanks
NUCLEAR FISSION AND FUSION
Nuclear fission
It is the disintegration of a heavy nucleus into two lighter nuclei.
Energy is released during nuclear fission
An example is the bombardment of uranium
U
235
92
by slow neutrons
energynKrBaUnU
1
0
92
36
141
56
236
92
1
0
235
92
3
Nuclear fission gives out a huge amount of heat which is used to turn water into steam.
The steam is used in turn to drive turbans for electricity
The energy released by the fission of a single uranium atom is about 200 Mev, and about
80% of this goes into providing the kinetic energy of the two fission fragments which are
often radioactive.
Nuclear reactors makes use of controlled fission reaction to provide heat at a steady rate
The atomic bomb makes use of uncontrolled fission reaction
The energy from nuclear reactors is used to make steam for the turban in conventional
power station.
Nuclear fusion
This is the amalgamation (combination)of two light nuclei to produce a heavier nucleus
Like in fission, energy is released.
An example is the fusion of two deuterium nuclei to produce a helium-3
energynHeHH
1
0
3
2
2
1
2
1
Reactions of this type( the conversion of hydrogen to helium) are the sources of the sun’s
energy
The explosion of a hydrogen bomb is as a result of a nuclear fusion.
COMPARISON BETWEEN NUCLEAR FISSION AND NUCLEAR FUSION
NUCLEAR FISSION
NUCLEAR FUSION
1
It is the breaking of a heavy nucleus
to produce lighter ones
It is the amalgamation (combination)of two light
nuclei to produce a heavier nucleus
2
It is a chain process
It is not a chain process
184
3
It results into production of many
radioactive particles
It results into production of few radioactive
particles
4
Less energy is needed to activate
the process
High energy is needed to kick start the process
5
Releases fewer energy
Releases higher amounts of energy
6
Does NOT normally occur in nature
It normally occurs in nature ( like in the stars e.g.
sun
7
Its progress can be controlled
It’s very difficult to control its progress
8
Leaves many radioactive wastes
( causes pollution)
Leaves no trace of radioactive wastes hence NO
pollution
USES OF NUCLEAR RADIATION (RADIOACTIVITY)
1. Carbon 14 dating
It is used in archaeology to estimate age of an organism that died sometime
back.
Living matter takes in carbon in form of CO
2
from the atmosphere. When an
organism dies the intake of CO
2
stops. This causes carbon-14 content to start
decreasing as a result of radioactivity. However, the caron-12 content stays
constant and therefore from the moment of death the rate of carbon-14 to
carbon-12 decreases
When a tree is cut down and burnt radioactive decay of charcoal begins. The
longer the charcoal was formed the lower is its radioactivity.
2. Tracers
In agriculture it is used to study uptake of fertilizer in plants. E.g.
radiophosphosrous-32
Radioactive carbon-14 is also used in the study of photosynthesis.
Detecting leaks in underground pipes by adding a tracer to the fluid in the pipe.
Eg radioactive iodine-131 is used to detect leaks in water and fuel lines.
Note that tracers with short half-lives are used to avoid accumulation of the
radioactive substances which can pollute the environment and causes different
undesirable effects. It also assists in getting instant (immediate) results.
3. Thickness monitoring
The thickness of metal sheet, paper, rubber and plastics can be monitored during
manufacturing by passing it between beta or gamma source and a suitable
detector
The thicker the sheet the grater the absorption
Also in the industries they use radiation to check whether packets and
containers have been correctly filled.
4. Medical (radiotherapy)
185
Cancer cells can be destroyed by gamma radiation. Gamma rays are most
penetrating
Deep lying tumors can be treated by planting radium-226 or ceasium-137 inside
the body close to the tumor
Radioactive iodine 131 is used in persons suffering from diseases of the thyroid
gland. It may also be injected into the blood stream to detect tumors
Iron -59 is used to monitor red blood cells production in bone marrow
Sodium-24 is used to monitor blood flow and detect blood clots and destruction
in blood vessels.
5. Sterilization/food preservation
Gamma rays can be used to kill bacteria on such thing as hospital blankets and
certain foods. Surgical (medical) equipment can be sterilized more effectively by
radioactivity than boiling.
6. Power generation ( Energy source)
Power is produced at nuclear reactors. The obvious example is the use of
uranium-235 in fuel nuclear power stations to produce energy. This energy
source is called nuclear energy.
There are, however, other portable power sources which use radioactive
materials.
Some satellites contain radioactive material which are decaying naturally and
giving enough energy to remain hot throughout long space flight.
The great merit in this type of power station is the small fuel consumption. A few
kg of uranium are sufficient to produce hundreds of megawatts of power for a
whole day.
NUCLAR REACTORS
A nuclear reactor produces and controls the release of energy from splitting of atoms of certain
elements.
Energy released is used as heat to produce steam which drives turbines.
Some Major Components of a nuclear reactor:
Fuel
Uranium is the basic fuel. Usually pellets of uranium oxide (UO
2
) are arranged in tubes to form
fuel rods. The rods are arranged into fuel assemblies in the reactor core.
Moderator
This is material in the core which slows down the neutrons released from fission so that they
cause more fission. It is usually water, but may be heavy water or graphite.
Control rods
186
These are made with neutron-absorbing material such as cadmium, hafnium or boron, and are
inserted or withdrawn from the core to control the rate of reaction.
Coolant.
A liquid or gas that circulates through the core so as to transfer the heat from it. . In light water
reactors the water moderator functions also as primary coolant.
2015 No
6c
i. Mention any two properties of beta particles
Answer: are negatively charged, are more penetrating than alpha, are less ionizing
than alpha, get deflected in magnetic and electric fields etc
ii. State any two safety precautions when handling radioactive substances
Answer: handle with forceps, use minute amounts when doing experiments, keep it
away from eye and skin
iii. Explain how radioactivity could be induced
Answer: it could be induced by exposing radioactive nuclei to a fast moving particle eg
a neutron, alpha etc. The accelerated particle excites the nucleus to undergo
radioactivity.
d. figure 4 is diagram showing the effect of magnetic field on nuclear radiation
187
Identify the type of radiation P, Q and R
Answers: P: is alpha radiation Q is gamma radiation and R is beta radiation
2014
7
a. The equation below shows nuclear decay of carbon
C
14
6
and cobalt
Co
60
27
___________.2
__________.1
60
27
60
27
14
7
14
6
CoCo
NC
(i) Complete the equations
0
0
60
27
60
27
14
7
0
1
14
6
.2
.1
CoCo
NeC
(ii) Which equation represents gamma emission? Answer: Equation 2
(iii) Why has
Co
60
27
not changed its atomic mass and atomic number in the daughter
nuclide?
Answer: it’s because gamma emission is not a particle but form of energy with
zero atomic number and mass number hence no change to
Co
60
27
b. State any three properties of a beta particle
Answer:
Have a negative charge
Gets deflected in magnetic and electric field
Are more penetrating than alpha
Have a less ionization power than alpha
Move at relatively higher speed than alpha
c. What do A and Z stand for in the following nuclear notation?
X
A
Z
188
A: atomic mass or mass number
Z: atomic number
d. The half-life of sodium-24 is 15hours. If the initial count rate is 240
counts/second, calculate the count rate after 30 hours
sec/60
4
1
240
2
1
240
2
1
30sec,/240,15
15
30
0
0
countsNN
hrsTcNhrst
t
T
2013
7.
a. State any two safety measures when handling radioactive substances at a school
Use small amounts when conducting experiments
Handle with forceps
Avoid bringing samples close to one’s mouth and eye
Keep them away from food stuffs
b. Mention two particles that are produced when a neutron disintegrates
Answer: Protons and electrons
c. Explain how gamma emission occurs
Gamma emissions are simply packets of energy released by decay of atomic nuclei
as they move from a high state of instability (with high energy) to a low level of
stability (less energy).
d. A radioactive substance with an activity of 30 counts/second is disintegrating.
Calculate the number of disintegrated atoms after 10 minutes
2012 6d
(i) Name two types of radioactivity. Natural radioactivity and artificial (induced)
radioactivity
(ii) Define “half-life” of a radioactive element
This is the time taken for half of the radioactive element to decay
(iii) Mention any three properties of alpha particle
Time = 10 minutes = 60 x 10 = 600 seconds
If in 1 second, 30 atoms disintegrates then in 600 seconds it
would be more
Number of disintegrated atoms = 600 x 30 = 18 000
189
1. It is the most highly ionizing particle. I.e. it knocks off electrons in an atom and
makes it charged. This is attributed to the high charge they have ( +2)
2. It can penetrate about 5 cm of air because of its relatively large mass.
3. It can be deflected by electric and magnetic fields. (It bends in these fields). This is
so because of its positive charge and any charge creates a magnetic and an electric
field.
4. It is positively charged( has a charge of +2)
5. Move at a low speed as compared to the speed of light (due to its large mass).
8d .Explain any two uses of nuclear radiation. Source of energy. This is done in nuclear
reactors. Uranium is normally used. Used in sterilization of medical equipment. Gamma
rays are best used because of high penetration.
PRACTICE QUESTIONS (ANSWERS TO SELECTED QUESTIONS ARE PROVIDED)
1. The following represents the radioactive decay of throrium-232. A, Z and X are unknown.
90
232
Th
Z
A
X +
2
4
a. What type of radiation is being emitted?
b. What are the values of A and Z?
c. Use the periodic table to determine what X is
d. Rewrite the above equation replacing A, Z and X with numbers and symbols you have
found.
e. What are the decay products?
2. When a radioactive sodium-24 decays, magnesium -24 is formed. The following is an incomplete
equation for the process.
11
24
Na
12
24
Mg + ______________
Assuming that only one charged particle is produced
a. What is the mass number of this particle?
b. What is the relative charge of this particle?
c. What type of particle is it?
3. The half-life of Iodine-128 is 25 minutes. If the activity of a sample of iodine-128 is
800becquerels (800 Bq). What would you expect the activity to be after
a. 25 minutes
b. 50 minutes
c. 100 minutes
4. Give two uses of radioactive tracers and explain why it is important to use tracers with short
half-lives.
5. The symbol
17
35
Cl represents one atom of Chlorine
State the names, numbers and locality of the different particles found in one of these chlorine
atoms.
190
6. The phosphorous-32 is a radioactive isotope. It can be used to prove that plants absorb
phosphorous from the soil around them.
a.
I. The stable isotope of phosphorous has a mass number of 31. Sate the structural difference
between the atoms of phosphorous-31 and phosphorous-32.
II. Explain why both isotopes of phosphorous have identical chemical properties.
b. Phosphosrous-32 is a beta-emitter with a half-life of 14 days.
I. What is a beta emitter?
II. The atomic number of phosphosrous-32 atom is 15. State the new values of the atomic and
mass numbers of the atom just after it has emitted a beta particle.
c. State three safety precautions which should be adopted when doing experiments with radioactive
phosphorous-32.
7. A milk sample containing Iodine-131 was found to have an activity of 1600 units per liter. The
activity of the sample was measured every 7 days and the results are as shown below
TIME ( days)
0
7
14
21
28
35
ACTIVITY ( units per day)
1600
875
470
260
140
77
a. Draw a graph of activity against time.
b. Estimate the half-life of iodine 131 and show on the graph how you arrive at your
answer.
8. Fig below shows how radon decays.
(Identify the values of a, b, c, d, e, f, g, h, I and j
9. Write down the nuclear equation for the following fro the following
a. Conversion of
6
13
C to
6
14
C
b. Conversion of
15
30
P to
14
30
Si
10. A radioactive series is shown below. Copy and fill the missing words and numbers.
80
210
A
a
78
b
B
c
d
206
C
alpha
f
e
D
beta
h
g
E
i
j
198
F
Note: there are 10 question marks for you to answer.
191
11. Bismuth sample with a half-life of 5 days was left to disintegrate. It was observed using a
Geiger Muller Counter that after 25 days from the start of decay the activity of Bismuth
sample was 10counts/second. Calculate the original counts of the Bismuth sample.
.
12. Complete the following nuclear equations
a. __________ +
2
4
He
15
30
P +
0
1
n
b. __________ +
1
2
H
7
13
N +
0
1
n
c. __________
36
82
Kr +
1
0
e
d.
29
66
Cu
________+
1
0
e
e.
1
0
e + _______
3
7
Li
13. Write four properties of each of the following
a) Alpha decay
b) Beta decay
c) Gamma radiation
14. By giving an example of each in form of equation distinguish nuclear fusion from nuclear
fission
15. Highlight six uses of nuclear radiation or radioactivity.
16. Mention five precautionary measures of handling radioactive materials.
Write down four ways of detecting nuclear radiation
ANSWERS TO SELECTED QUESTIONS
1.
a. Alpha radiation
b. A= 238 -4 = 228 and Z = 90-2 = 88
c. X is Radium
d.
4
2
228
88
232
90
RaTh
e. Alpha particles and radium
2.
a. 0
b. -1
c. A beta particle
3.
192
a. 400Bq
b. 200Bq
c. 50Bq
4. To detect uptake of nutrients by the plants and to detect cracks in pipes. It is important
to use tracers with short half-lives because it assists in avoiding accumulation and over
stay of the substances in the environment which can be hazardous to the living things. It
also assists in getting quick results.
5.
Name of particle
locality
number
Electron
Shells/energy levels
17
Proton
nucleus
17
neutron
nucleus
18
6.
a. (i)The atoms will defer in the number of neutrons that will be contained in their
nucleus.
(ii) The isotopes have identical chemical properties due to same number of
electrons and protons.
b. (i) It is a substance that releases a beta particle when it undergoes radioactive
decay.
(ii) Mass number will be 32 and the atomic number will become 16
c. Refer to the notes.
7a. THE GRAPH OF ACTIVITY AGAINST TIME
0
200
400
600
800
1000
1200
1400
1600
1800
0 10 20 30 40
Activity (Units per day)
Time (days)
193
7. b. 8 days
8.
a
b
c
d
e
f
g
h
i
j
86
218
214
82
214
83
214
84
210
82
9.
a.
CnC
14
6
1
0
13
6
b.
SieP
30
14
0
1
30
15
10.
a
b
c
d
e
f
g
h
i
j
Alpha
206
Beta
79
202
77
202
78
Alpha
76
11. Use the table below
Time (days)
25
20
15
10
5
0
Amount remaining
10cts/sec
20cts/sc
40cts/sec
80cts/sec
160cts/sec
320cts/sec
Therefore the original count was 320 counts per second
Or using the formula as shown on the next page
5
0
5
25
00
2
1
2
1
/10
2
1
NNscNN
t
T
Thus
sec/320/1032
32
1
/10
00
countsscxNNsc
12.
a
b
c
d
e
Al
27
13
C
12
6
rB
82
35
Zn
66
30
Be
7
4
194
REFERENCES:
1. Wallis, K (1992) Physical science for Malawi Book 1 &2, Zomba; Chancellor College
2. Duncan, T. (1983). Physics for Today and Tomorrow, 2
nd
Ed. London: John Murray
3. Gallagher, R .M and Ingram, P. (1997). GCSE CHEMISTRY. Oxford: Oxford University Press.
4. Gallagher, R.M (2000), Complete Chemistry, Oxford: Oxford University Press.
5. Odian, G &Blei, I (1994) Schaum’s Outline of Theory & Problems of General, Organic, and
Biological Chemistry: Mac Graw Hill Companies: New York
6. MIE (2001), Malawi Senior Secondary Teaching Syllabus, Physical Science (3-4),Zomba: MIE
7. Malawi National Examinations Board, Maneb- Physical Science Examination Past papers.
8. Murigi, M. etal (2007)K K.C.S.E Golden Tips Physics, Moran Publishers
